Quantities in Chemical Reactions

Transcription

1 Chapter Quantities in Chemical Reactions Note to teacher: You will notice that there are two different formats for the Sample Problems in the student textbook. Where appropriate, the Sample Problem contains the full set of steps: Problem, What Is Required, What Is Given,, Act on Your Strategy, and. Where a shorter solution is appropriate, the Sample Problem contains only two steps: Problem and Solution. Where relevant, a Check Your Solution step is also included in the shorter Sample Problems. Solutions for Practice Problems Student Textbook page. Problem Consider the following reaction. H (g) + O (g) H O (l) (a) Write the ratio of H molecules: O molecules: H O molecules. (b) How many molecules of O are required to react with 00 molecules of H, according to your ratio in part (a)? (c) How many molecules of water are formed when 78 molecules of O react with H? (d) How many molecules of H are required to react completely with molecules of O? (a) You need to give the ratio of molecules in the reaction. (b) You need to find the number of O molecules that will react with the number of H molecules. (c) You need to find the number of water molecules that will form from the number of reactants. (d) You need to find the number of H molecules that will react with the number of O molecules. The balanced equation for the reaction is. The number of molecules of reactants or products is in each question. (a) The ratio is by the whole number in front of each reactant and product in the balanced equation. (b) Equate the ratio of number of molecules to the whole number ratio obtained in (a) for hydrogen:oxygen, and solve for the number of oxygen molecules. (c) As in (b), equate the ratio of the number of molecules to the whole number ratio obtained in (a) for oxygen:water. Solve for the number of water molecules. (d) As in (b) and (c), equate the ratio of the number of molecules to the whole number ratio obtained in (a) for hydrogen:oxygen. Solve for the number of hydrogen molecules. 6

2 H (g) + O (g) H O (l) (a) (b) (c) (d) 00 molecules 50 molecules 78 molecules molecules 956 molecules.0 0 molecules (a) The ratio is : : molecules H (b) = N molecules O molecule O 00 molecules H N = 50 molecules O (c) (d) molecule O molecules H O = 78 molecules O N molecules H O N = 956 molecules H O molecules H = N molecules H molecule O N =.0 0 molecules H (a) Try drawing the molecules of hydrogen and oxygen and modelling what happens when they react to form water molecules. You will see that it takes full molecules of gaseous H and one full molecule of gaseous O to form full molecules of water. (b) The ratio of hydrogen to oxygen in this reaction is always :. Dividing both 00 molecules of hydrogen and 50 molecules of oxygen by 50 will give the lowest ratio of :. Your answer is consistent. (c) The ratio of oxygen to water in this reaction is always :. Dividing both the 78 molecules of oxygen and 956 molecules of water by 78 will give the lowest ratio of :. Your answer is consistent. (d) The ratio of hydrogen to oxygen in this reaction is always :. Dividing both.0 0 molecules of hydrogen and molecules of oxygen by will give the lowest ratio of :. Your answer is consistent.. Problem Iron reacts with chlorine gas to form iron(iii) chloride, FeCl. Fe (s) + Cl (g) FeCl (s) (a) How many atoms of Fe are needed to react with three molecules of Cl? (b) How many molecules of FeCl are formed when 50 atoms of Fe react with sufficient Cl? (c) How many Cl molecules are needed to react with.0 0 atoms of Fe? (d) How many molecules of FeCl are formed when molecules of Cl react with sufficient Fe? (a) You need to give the ratio of Fe to Cl molecules in the reaction. (b) You need to find the number of FeCl molecules that will form from the number of Fe atoms. (c) You need to find the number of Cl molecules that will react with the number of Fe. (d) You need to find the number of FeCl molecules that will form from the number of Cl molecules. 65

3 The balanced equation for the reaction is. The number of molecules of reactants or products is in each question. (a) The ratio is by the whole number in front of each reactant and product in the balanced equation. (b) Equate the ratio of number of molecules to the whole number ratio for Fe:FeCl, and solve for the number of FeCl molecules. (c) As in (b), equate the ratio of the number of molecules to the whole number ratio for Fe:Cl. Solve for the number of Cl molecules. (d) As in (b) and (c), equate the ratio of the number of molecules to the whole number ratio for Cl :FeCl. Solve for the number of FeCl molecules. (a) (b) (c) (d) Fe (s) atoms 50 atoms.0 0 atoms + Cl (g) molecules molecules molecules FeCl (s) 50 molecules.0 0 molecules (a) According to the balanced equation, only two atoms of Fe are required to react with three molecules of Cl. (b) atoms of Fe = formula units of FeCl 50 atoms of Fe N formula units of FeCl N = 50 formula units of FeCl atoms of Fe (c) =.0 0 atoms of Fe molecules of Cl N molecules of Cl N = molecules of Cl (d) molecules of Cl molecules of FeCl = molecules of Cl N molecules of FeCl N =.0 0 molecules of FeCl (a) Try drawing the atoms and molecules of iron and chloride gas and modelling what happens when they react to form iron(iii) chloride molecules. You will see that it takes atoms of Fe and one molecules of gaseous Cl to form full molecules of FeCl. (b) The ratio of iron to iron(iii) chloride in this reaction is always :, which is : in the lowest ratio. Dividing both the 50 atoms of Fe and 50 molecules of FeCl by 50 will give the lowest ratio of :. Your answer is consistent. (c) The ratio of Fe to Cl in this reaction is always :, which is :.5 in the lowest ratio. Dividing both the.0 0 atoms of Fe and the molecules of Cl by.0 0 will give the lowest ratio of :.5. Your answer is consistent. (d) The ratio of Cl to FeCl in this reaction is always :, which is.5: in the lowest ratio. Dividing both molecules of Cl and.0 0 molecules of FeCl by.0 0 will give the lowest ratio of.5:. Your answer is consistent. 66

4 . Problem Consider the following reaction. Ca(OH) (aq) + HCl (aq) CaCl + H O (l) (a) How many formula units of calcium chloride, CaCl, would be produced by molecules of hydrochloric acid, HCl? (b) How many molecules of water would be produced in the reaction in part (a)? (a) You need to find the number of CaCl formula units produced by the amount of HCl. (b) You need to find the number of water molecules produced by the amount of HCl. The balanced equation for the reaction is. The number of molecules of HCl is. (a) Equate the ratio of number of molecules to the whole number ratio for HCl:CaCl and solve for the number of CaCl formula units. (b) As in (a), equate the ratio of the number of molecules to the whole number ratio for HCl:H O. Solve for the number of H O molecules. Ca(OH) (aq) + HCl (aq) CaCl (g) + H O (l) molecule or formula unit ratio (a) (b) N N (a) N = ( ) =. 0 formula units of CaCl 5 (b) N = ( ) = molecules H. 0 O (a) The of HCl to CaCl in this reaction is always :. Dividing both the molecules of HCl and molecules of water by will give the lowest ratio of :. Your answer is consistent (b) The of HCl to water in this reaction is always :, which is : in the lowest ratio. Dividing both the of HCl and molecules of water by will give the lowest ratio of :. Your answer is consistent. Solutions for Practice Problems Student Textbook page 5. Problem Aluminum bromide can be prepared by reacting small pieces of aluminum foil with liquid bromine at room temperature. The reaction is accompanied by flashes of red light. Al (s) + Br (l) AlBr (s) How many moles of Br are needed to produce 5 mol of AlBr, if sufficient Al is present? 67

5 You need to find the number of moles of Br to produce the number of moles of product. The balanced equation for the reaction is. The number of moles of product is. From the balanced equation, determine the mol ratio of reactants to products, by the whole number in front of each molecule. Equate the equation mol ratio of Br :AlBr to the same ratio using the number of mol of AlBr, and solve for the number of moles of Br. Al (s) + Br (l) AlBr (s) 5 mol n molbr molalbr = n mol Br 5molAlBr n = 5 mol AlBr molbr molalbr = 7.5 mol Br The mol ratio of Br to AlBr in this reaction is always :, which is.5: in the lowest ratio. Dividing both the 7.5 mol of Br and 5.0 mol of AlBr by 5.0 will give the lowest ratio of.5:. Your answer is consistent. 5. Problem Hydrogen cyanide gas, HCN (g), is used to prepare clear, hard plastics, such as Plexiglas. Hydrogen cyanide is formed by reacting ammonia, NH, with oxygen and methane, CH. NH (g) + O (g) + CH (g) HCN (g) + 6H O (g) (a) How many moles of O are needed to react with. mol of NH? (b) How many moles of H O can be expected from the reaction of.5 mol of CH? Assume that sufficient NH and O are present. (a) You need to find the number of moles of O that will react with the number of moles of ammonia. (b) You need to find the number of moles of water that will be produced from the number of moles of methane. The balanced equation for the reaction is. The number of moles of reactants is. (a) From the balanced equation, determine the mol ratio of reactants to products, by the whole number in front of each molecule. Equate the equation mol ratio of NH :O to the same ratio using the number of mol of NH, and solve for O. (b) As in (a), equate the ratio of the number of moles to the whole number ratio for CH :H O. Solve for the number of moles of H O. 68

6 NH (g) + O (g) CH (g) HCN (g) + 6H O (g) Mole ratio 6. mol n.5 mol n (a) (b) molnh =. mol NH molo n mol O n =. mol NH molo =.8 mol O molnh molch 6molH O =.5 mol CH n mol H O n =.5 mol CH 6molH O molch = 7.5 mol H O (a) The mol ratio of NH :O in this reaction is always :, the lowest ratio of which is :.5. Dividing both the. mol of NH and.8 mol of O by. will give the lowest ratio of.5:. Your answer is consistent. (b) The mol ratio of CH :water in this reaction is always :6, which is : in the lowest ratio. Dividing both the.5 mol of CH and 7.5 mol of water by.5 will give the lowest ratio of :. Your answer is consistent. 6. Problem Ethane gas, C H 6, is present in small amounts in natural gas. It undergoes complete combustion to produce carbon dioxide and water. C H 6(g) + 7O (g) CO (g) + 6H O (l) (a) How many moles of O are required to react with.9 mol of C H 6? (b) How many moles of H O would be produced by.0 mol of O and sufficient ethane? (a) You need to find the number of moles of O that will react with the number of moles of ethane. (b) You need to find the number of moles of water that will be produced from the number of moles of oxygen. The balanced equation for the reaction is. The number of moles of reactants is. (a) From the balanced equation, determine the mol ratio of reactants to products, by the whole number in front of each molecule. Equate the equation mol ratio of C H 6 :O to the same ratio using the number of mol of ethane, and solve for O. (b) As in (a), equate the ratio of the number of moles to the whole number ratio for O :H O. Solve for the number of moles of H O. C H 6(g) + 7O (g) CO (g) + 6H O (g) mol n.0 mol n 69

7 (a) (b) molc H 6 7molO =.9 mol C H 6 n mol O n =.9 mol C H 6 7molO 7molO 6molH O =. mol O n mol O n =. mol O molc H 6 = 8.7 mol O 6molH O 7molO =. mol H O (a) The mol ratio of C H 6 :O in this reaction is always :7, the lowest ratio of which is :.5. Dividing both the.9 mol of C H 6 and 8.7 mol of O by.9 will give the lowest ratio of :.5. Your answer is consistent. (b) The mol ratio of O :water in this reaction is always 7:6, which is.67: in the lowest ratio. Dividing both the.0 mol of O and.0 mol of water by.0 will give the lowest ratio of.67:. Your answer is consistent. 7. Problem Magnesium nitride reacts with water to produce magnesium hydroxide and ammonia gas, NH according to the balanced chemical equation Mg N (s) + 6H O (l) Mg(OH) (s) + NH (g) (a) How many molecules of water are required to react with. mol Mg N? (b) How many molecules of Mg(OH) will be expected in part (a)? (a) You need to find the number of molecules of water that will react with the magnesium nitride. (b) You need to find the number of molecules of magnesium hydroxide that will form from the amount of magnesium nitride. The balanced equation for the reaction is. The number of moles of magnesium nitride is. Avogadro s number, N A = molecules/mol. (a) From the balanced equation, determine the mol ratio of reactants to products, by the whole number in front of each molecule. Equate the equation mol ratio of Mg N :H O to the same ratio using the number of mol of magnesium nitride, and solve for the number of moles of O. Multiply this by Avogadro s number to obtain the number of molecules needed. (b) As in (a), equate the ratio of the number of moles to the whole number ratio for Mg N :Mg(OH). Solve for the number of moles of Mg(OH). Multiply this by Avogadro s number to obtain the number of molecules produced. Mg N (s) + 6H O (l) Mg(OH) (s) + NH (g). mol 6 n n (a) (b) molmg N 6molH O =. mol Mg N n mol H O n =. mol Mg N 6molH O molmg N =.8 mol H O N =.8 mol H O (6.0 0 ) molecules/mol = 8. 0 molecules H O molmg N molmg(oh) =. mol Mg N n mol Mg(OH) n =. mol Mg N molmg(oh) molmg N = 6.9 mol Mg(OH) 70

8 N = 6.9 mol (6.0 0 ) molecules/mol =. 0 molecules Mg(OH) (a) The mol ratio of Mg N :H O in this reaction is always :6. Dividing both the. mol of Mg N and.8 mol of H O by. will give the lowest ratio of :6. Your answer is consistent. (b) The mol ratio of Mg N :Mg(OH) in this reaction is always :. Dividing both the. mol of Mg N and 6.9 mol of Mg(OH) by. will give the lowest ratio of :. Your answer is consistent. Solutions for Practice Problems Student Textbook page 7 8. Problem Vanadium can form several different compounds with oxygen, including V O 5, VO, and V O. (a) How many moles of V are needed to produce 7.7 mol of VO? Assume that sufficient O is present and that V and O react to form VO only. (b) How many moles of V are needed to react with 5.9 mol of O to produce V O? Assume that V and O react to form V O only. (a) You need to find the number of moles of V to produce the amount of VO. (b) You need to find the number of moles of V that will react with the amount of oxygen to form V O. The number of moles of product or reactant is. The type of reactants and products that make up each reaction are. (a) Write out the full reaction and balance the equation. Determine the of reactant to products, which is the whole number in front of each reactant and product. Equate the ratio of V:VO in the equation to the amounts, and solve for the number of moles of V. (b) As in (a), write out the full reaction and balance the equation. Determine the of reactant to products. Equate the ratio of V:O in the equation to the amounts, and solve for the number of moles of V. (a) V (s) + O (g) VO (s) 7.7 mol n molv = molvo n = 7.7 mol VO n mol V 7.7 mol VO molv molvo = 7.7 mol V 7

9 (b) V (s) + O (g) V O (s) 5.9 mol n molv = molo n mol V 5.9 mol O n = 5.9 mol O molv molo = 7.9 mol V (a) The mol ratio of V:VO in this reaction is always :. Dividing both the 7.7 mol of V and 7.7 mol of VO by 7.7 will give the lowest ratio of :. Your answer is consistent. (b) The mol ratio of V:O in this reaction is always :, which is.: in the lowest ratio. Dividing both the 7.9 mol of V and 5.9 mol of O by 5.9 will give the lowest ratio of.: Your answer is consistent. 9. Problem Nitrogen, N, can combine with oxygen, O, to form several different oxides of nitrogen. These oxides include NO and N O. N (g) +O (g) Æ N O (g) N (g) +O (g) Æ NO (a) How many moles of O are required to react with mol of N to form N O? (b) How many moles of O are required to react with mol of N to form NO? (a) You need to find the number of moles of oxygen needed to produce N O from the amount of nitrogen. (b) You need to find the number of moles of oxygen needed to produce NO from the amount of nitrogen. The number of moles of nitrogen is. The type of reactants and products that make up each reaction are. (a) Write out the full reaction and balance the equation. Determine the of reactant to products, which is the whole number in front of each reactant and product. Equate the ratio of N :O in the equation to the amounts, and solve for the number of moles of O. (b) As in (a), write out the full reaction and balance the equation. Determine the of reactant to products. Equate the ratio of N :O in the equation to the amounts, and solve for the number of moles of O. (a) N (g) + O (g) N O (g) (9.5)(0 - ) mol n 7

10 moln molo = mol N n mol O - mol O - mol N n = ( ) mol N = mol O (b) N (g) + O (g) NO (g) (9.5 )(0 - ) mol n moln = mol N molo n mol O - mol O mol N n = mol N = mol O (a) The mol ratio of N :O in this reaction is always :. Dividing both the mol of N and mol of O by will give the lowest ratio of :. Your answer is consistent. (b) The mol ratio of N :O in this reaction is always :. Dividing both the mol of N and 0.9 mol of O by will give the lowest ratio of : Your answer is consistent. 0. Problem When heated in a nickel vessel to 00 C, xenon can be made to react with fluorine to produce colourless crystals of xenon tetrafluoride according to the following equation Xe (g) + F (g) Æ XeF (g). (a) How many moles of fluorine gas, F, would be required to react with mol of xenon? (b) Under somewhat similar reaction conditions, xenon hexafluoride can also be obtained. How many moles of fluorine would be required to react with the amount of xenon in part (a) to produce xenon hexafluoride? (a) You need to find the number of moles of F that will react with the amount of xenon to form xenon tetrafluoride. (b) You need to find the number of moles of F that will react with the amount of xenon to form xenon hexafluoride. The number of moles of xenon is. The type of reactants and products that make up each reaction are. (a) Write out the full reaction and balance the equation. Determine the of reactant to products, which is the whole number in front of each reactant and product. Equate the ratio of Xe:F in the equation to the amounts, and solve for the number of moles of F. (b) As in (a), write out the full reaction and balance the equation. Determine the of reactant to products. Equate the ratio of Xe:F in the equation to the amounts, and solve for the number of moles of F. 7

11 (a) Xe (g) + F (g) XeF (s) mol n (b) molxe =.5 0- mol Xe molf n mol F n = ( ) mol Xe molf molxe = mol F Xe (g) + F (g) XeF 6(s) mol n molxe =.5 0- mol Xe molf n mol F n = ( ) mol Xe molf molxe =.06 mol F (a) The mol ratio of Xe:F in this reaction is always :. Dividing both the 0.5 mol of Xe and mol of F by 0.5 will give the lowest ratio of :. Your answer is consistent. (b) The mol ratio of Xe:F in this reaction is always :. Dividing both the 0.5 mol of Xe and.06 mol of F by 0.5 will give the lowest ratio of :. Your answer is consistent. Solutions for Practice Problems Student Textbook page. Problem Ammonium sulfate, (NH ) SO, is used as a source of nitrogen in some fertilizers. It reacts with sodium hydroxide to produce sodium sulfate, water, and ammonia. (NH ) SO (s) + NaOH (aq) Na SO (aq) + NH (g) + H O (l) What mass of sodium hydroxide is required to react completely with 5. g of (NH ) SO? You need to find the mass of NaOH that will complete the reaction with the amount of ammonium sulfate. The mass of the ammonium sulfate is. The balanced equation is. Step Convert the mass (m) of ammonium sulfate to the number of moles (n) of ammonium sulfate, using its (M). Use the formula n = m / M Step Determine the of NaOH to (NH ) SO from the balanced equation. Using the result in Step, solve for the number of moles of NaOH. Step Convert the number of moles of NaOH to mass using its. Use the formula m = n M of the NaOH. 7

12 (NH ) SO (s) + NaOH (aq) Na SO (aq) + NH (g) + H O (l).7 g/mol 0.0 g/mol.05 g/mol 7.0 g/mol 8.0 g/mol 5. g m Step 5. g (NH ) SO n = = 0.6 mol (NH. 7g/mol ) SO mol (NH Step ) SO 0. 6 mol (NH ) SO = mol NaOH n NaOH mol NaOH n = mol (NH ) SO = 0. mol NaOH mol (NH ) SO Step m = 0. mol NaOH 0.0 g/mol = 9.8 g NaOH Therefore, the mass of NaOH that will react completely with the ammonium sulfate is 9.8 g. The mol ratio of (NH ) SO :NaOH in this reaction is always :. Dividing both the 0.6 mol of (NH ) SO and 0. mol of NaOH by 0.6 will give the lowest ratio of :. Your answer is consistent.. Problem Iron(III) oxide, also known as rust, can be removed from iron by reacting it with hydrochloric acid to produce iron(iii) chloride and water. Fe O (s) + 6HCl (aq) FeCl (aq) + H O (l) What mass of hydrogen chloride is required to react with.00 0 g of rust? You need to find the mass of HCl that will complete the reaction with the amount of rust. The mass of the HCl is. The balanced equation is. Step Convert the mass (m) of rust to the number of moles (n) of rust, using its (M). Use the formula n = m / M Step Determine the of rust to HCl from the balanced equation. Using the result in Step, solve for the number of moles of HCl. Step Convert the number of moles of HCl to mass using its. Use the formula m = n M of the HCl. Fe O (s) + 6HCl (aq) FeCl (aq) + H O (l) g/mol 6.6 g/mol 6. g/mol 8.0 g/mol (.00)(0 ) g m Step n = (.00)(0 ) gfe O = 0.66 mol Fe 59.7 g/mol O Step molfe O = 0.66 mol Fe O 6molHCl n mol HCl n = mol HCl. mol Fe O =. 76 mol HCl mol Fe O Step m =.76 mol HCl 6.6 g/mol = 7 g HCl Therefore, the mass of HCl needed to bring this reaction to completion is 7 g. 75

13 The mol ratio of Fe O :HCl in this reaction is always :6. Dividing both the 0.66 mol of rust and.76 mol of HCl by 0.66 will give the lowest ratio of :6. Your answer is consistent.. Problem Iron reacts slowly with hydrochloric acid to produce iron(iii) chloride and hydrogen gas. Fe (s) + HCl (aq) FeCl (aq) + H (g) What mass of HCl is required to react with.56 g of iron? You need to find the mass of HCl that will complete the reaction with the amount of iron. The mass of the iron is. The balanced equation is. Step Convert the mass (m) of iron to the number of moles (n)of iron, using its (M). Use the formula n = m / M Step Determine the of Fe to HCl from the balanced equation. Using the result in Step, solve for the number of moles of HCl. Step Convert the number of moles of HCl to mass using its. Use the formula m = n M of the HCl. Fe (s) + HCl (aq) FeCl (aq) + H (g) g/mol 6.6 g/mol 6.75 g/mol.0 g/mol.56 g m.56 gfe g/mol Step n = = mol Fe Step molfe mol Fe = molhcl n mol HCl n = mol HCl. mol Fe = 0. 7 mol HCl mol Fe Step m = 0.7 mol HCl 6.6 g/mol =.6 g HCl Therefore, the mass of HCl needed to react fully with the Fe is.6 g. The mol ratio of Fe:HCl in this reaction is always :. Dividing both the mol of Fe and 0.7 mol of HCl by will give the lowest ratio of :. Your answer is consistent.. Problem Dinitrogen pentoxide is a white solid. When heated it decomposes to produce nitrogen dioxide and oxygen. N O 5(s) Æ NO (g) + O (g) What volume of oxygen gas at STP will be produced in this reaction when. g of NO are made? You need to find the volume of oxygen that will be produced at STP together with the amount of NO. 76

14 The mass of the NO product is. The balanced equation is. Step Convert the mass (m) of NO to the number of moles (n) of NO, using its (M). Use the formula n = m / M Step Determine the of NO to O from the balanced equation. Using the result in Step, solve for the number of moles of O. Step Convert the number of moles of O to volume using the molar volume. L/mol at STP. N O 5(s) NO (g) + O (g) 08.0 g/mol 6.0 g/mol.00 g/mol. g m Step Step. g NO mol NO 6.0g/mol mol NO mol NO = mol O n mol O mol O mol NO n = = n = mol O = mol O. L Step V = 0.07 mol O = 0. 8 L mol Therefore, the volume of O that will be produced is 0.8 L. The mol ratio of NO :O in this reaction is always :. Dividing both the mol of NO and 0.07 mol of O by 0.07 will give the lowest ratio of :. Your answer is consistent. Solutions for Practice Problems Student Textbook 5. Problem Powdered zinc reacts rapidly with powdered sulfur in a highly exothermic reaction. 8Zn (s) + S 8(s) 8ZnS (s) What mass of zinc sulfide is expected when.0 g of S 8 reacts with sufficient zinc? You need to find the mass of ZnS that will be produced from the amount of sulfur. The mass of the S 8 is. The balanced equation is. Step Convert the mass (m) of sulfur to the number of moles (n) of sulfur, using its (M). Use the formula n = m / M Step Determine the of sulfur to zinc sulfide from the balanced equation. Using the result in Step, solve for the number of moles of ZnS. Step Convert the number of moles of ZnS to mass using its. Use the formula m = n M of the ZnS. 77

15 8Zn (s) + S 8(s) 8ZnS (s) g/mol g/mol 97.6 g/mol.0 g m Step Step.0 g S 0.5 mol S g/mol 8 mol S mol S8 = 8 mol ZnS mol S8 8 mol ZnS mol S 8 n = = n = 0. 5 mol S8 =.00 mol ZnS g Step m =.00 mol ZnS 97.6 g/mol = 97.5 ZnS Therefore, the mass of ZnS produced is 97.5 g. The mol ratio of S 8 :ZnS in this reaction is always :8. Dividing both the 0.5 mol of S 8 and.00 mol of ZnS by 0.5 will give the lowest ratio of :8. Your answer is consistent. 6. Problem The addition of concentrated hydrochloric acid to manganese(iv) oxide leads to the production of chlorine gas. HCl (aq) + MnO (g) MnCl (aq) + Cl (g) + H O (l) What volume of chlorine at STP can be obtained when g of HCl react with sufficient MnO? You need to find the volume of Cl that will be produced at STP from the amount of HCl. The mass of the HCl is. The balanced equation is. Step Convert the mass (m) of HCl to the number of moles (n) of HCl, using its (M). Use the formula n = m/m Step Determine the of HCl to Cl from the balanced equation. Using the result in Step, solve for the number of moles of Cl. Step Convert the number of moles of Cl to volume using the molar volume. L/mol at STP. HCl (aq) + MnO (g) MnCl (aq) + Cl (g) + H O (l) 6.6 g/mol 86.9 g/mol 5.8 g/mol 70.9 g/mol 6.0 g/mol (.76)(0 - ) g m Step Step - (.76 0 ) g HCl 6. 6 g/mol mol HCl mol HCl = mol Cl n mol HCl n = = 0.00 mol HCl mol Cl mol HCl n = mol HCl =.8 0 mol Cl Step m =.8 0 mol Cl. L/mol = L 78

16 The mol ratio of HCl:Cl in this reaction is always :. Dividing both the 0.00 mol of HCl and mol of Cl by.8 0 will give the lowest ratio of :. Your answer is consistent. 7. Problem Aluminum carbide, Al C, is a yellow powder that reacts with water to produce aluminum hydroxide and methane. Al C (s) + H O (l) Al(OH) (s) + CH (g) What volume of methane gas at STP is produced when water reacts with 5.0 g of aluminum carbide? You need to find the volume of CH at STP that is needed to react with the amount of Al C. The mass of the Al C is. The balanced equation is. Step Convert the mass (m) of Al C to the number of moles (n) of Al C, using its (M). Use the formula n = m / M Step Determine the of Al C to H O from the balanced equation. Using the result in Step, solve for the number of moles of H O. Step Convert the number of moles of CH to volume using the molar volume. L/mol at STP. Al C (s) + H O (l) Al(OH) (s) + CH (g).95 g/mol 8.0 g/mol 78.0 g/mol 6.05 g/mol 5.0 g V Step Step 50 g 0.7 mol Al 95 g/mol C. mol Al C mol Al C = mol CH n mol CH mol CH mol Al C n = = n = 0 7. mol Al C = 0. 5 mol CH Step V=0.5 mol CH. L/mol =.7 L Therefore the volume of methane that will be produced with aluminum carbide is.7 L. The mol ratio of Al C :CH in this reaction is always :. Dividing both the 0.7 mol of Al C and 0.5 mol of CH by 0.7 will give the lowest ratio of :. The answer is consistent. 8. Problem Magnesium oxide reacts with phosphoric acid, H PO, to produce magnesium phosphate and water. MgO (s) + H PO (aq) Mg (PO ) (s) + H O (l) How many grams of magnesium oxide are required to react completely with.5 g of phosphoric acid? 79

17 You need to find the mass of MgO that is needed to react with the amount of phosphoric acid. The mass of the H PO is. The balanced equation is. Step Convert the mass (m) of H PO to the number of moles (n) of H PO, using its (M). Use the formula n = m / M Step Determine the of H PO to MgO from the balanced equation. Using the result in Step, solve for the number of moles of MgO. Step Convert the number of moles of MgO to mass using its. Use the formula m = n M of the MgO. MgO (s) + H PO (aq) Mg (PO ) (s) + H O (l) 0. g/mol m 98.0 g/mol.5 g Step n =.5 gh PO 98.0 g/mol Step molmgo = molh PO = 0. mol H PO n mol MgO 0. mol H PO n = 0. mol H PO molmgo molh PO = 0.5 mol MgO Step m = 0.5 mol MgO 0. g/mol = 0.7 g Therefore, the mass of MgO that will react completely with the phosphoric acid is 0.7 g. The mol ratio of MgO:H PO in this reaction is always :, which is the same as.5:. Dividing both the 0.5 mol of MgO and 0. mol of H PO by 0. will give.5:. Your answer is consistent. Solutions for Practice Problems Student Textbook, page Problem Nitrogen gas is produced in an automobile air bag. It is generated by the decomposition of sodium azide, NaN. NaN (s) N (g) + Na (s) (a) To inflate the air bag on the driver s side of a certain car, 80.0 g of N is required. What mass of NaN is needed to produce 80.0 g of N? (b) How many atoms of Na are produced when 80.0 g of N are generated in this reaction? (a) You need to find the mass of NaN needed to produce the required amount of nitrogen. (b) You need to find the number of Na atoms produced in this reaction amount in (a). 80

18 The mass of the nitrogen produced is. The balanced equation is. Avogadro s number, N A = formula units/mol. (a) To obtain the mass of NaN, perform the following steps below: Step Convert the mass (m) of N to the number of moles (n) of N, using its (M). Use the formula n = m / M Step Determine the of NaN to N from the balanced equation. Using the result in Step, solve for the number of moles of NaN. Step Convert the number of moles of NaN to mass using its. Use the formula m = n M of the NaN. (b) Repeat Step in (a) using the of N to Na, to determine the number of moles of Na that is produced. Multiply this by the Avogadro s number to obtain the number of atoms of Na. (a) (b) Step Step NaN (s) 65.0 g/mol m 80 g N.86 mol N 8. 0 g/mol mol N.86 mol N = mol CH n mol NaN mol NaN mol N n = = n = 86. mol N = 9. mol NaN Step m =.9 mol NaN 65.0 g/mol = g NaN. Therefore, the mass of NaN needed is g. mol N.86 mol N = mol Na n mol Na mol Na mol N n = 86. mol N = 9. mol Na N (g) 8.0 g/mol 80.0 g + Na (s).99 g/mol N =.9 mol Na (6.0 0 ) atoms/mol =.5 0 atoms Na. Therefore, the number of atoms of Na produced is.5 0 atoms. (a) The mol ratio of NaN :N in this reaction is always :, which is :.5 in the lowest ratio. Dividing both the.9 mol of NaN and.86 mol of N by.9 will give the lowest ratio of :.5. Your answer is consistent. (b) Likewise, the mol ratio of N :Na in this reaction is always :, which is.5: in the lowest ratio. Dividing both the.86 mol of N and.9 mol of Na by.9 will give the lowest ratio of.5:. Your answer is consistent. 0. Problem The reaction of iron(iii) oxide with powered aluminum is known as the thermite reaction. Al (s) + Fe O (s) Al O (s) + Fe (l) (a) Calculate the mass of aluminum oxide, Al O, that is produced when. 0 atoms of Al react with Fe O. (b) How many formula units of Fe O are needed to react with 0. g of Al? 8

19 (a) You need to find the mass of Al O produced from the number of atoms of Al. (b) You need to find the number of Fe O molecules needed to react with the amount of Al. (a) The number of atoms of Al is. The balanced equation is. Avogadro s number, N A = formula units/mol. (b) The mass of the aluminum is. The balanced equation is. Avogadro s number, N A = formula units/mol. (a) Use the following steps below: Step Divide the number of Al atoms by Avogadro s number to obtain the number of moles in the reaction. Step Using the ratio of Al to Al O from the balanced equation and the answer in Step, solve for the number of moles of Al O. Step Multiply the number of moles of Al O by its to obtain the mass in grams. (b) Use the following steps below: Step Convert the mass (m) of Al to the number of moles (n) of Al, using its (M). Use the formula n = m / M Step Determine the of Al to Fe O from the balanced equation. Using the result in Step, solve for the number of moles of Fe O. Step Multiply the number of moles of Fe O to the Avogadro number to obtain the number of molecules in the reaction. (a) Al (s) + Fe O (s) Al O (s) + Fe (l) 6.98 g/mol 59.7 g/mol 0.96 g/mol g/mol. 0 atoms m Step n = Step (. 0 ) atoms Al (6.0 0 ) atoms/mol molal.6 mol Al = molal O n mol Al O =.6 mol Al n =.6 mol Al molal O =.8 mol Al molal O Step m =.8 mol Al O 0.96 g/mol = 0 g Al O Therefore, the mass of aluminum oxide produced is 0 g. (b) Al (s) + Fe O (s) Al O (s) + Fe (l) 6.98 g/mol 59.7 g/mol 9.96 g/mol g/mol 0. g N 0. g Al - Step n = =.97 0 mol Al g/mol mol Al mol Al Step = mol Fe O n mol Fe O - mol Fe O - mol Al n =.97 0 mol Al =. 8 0 Fe O 8

20 Step N = mol Fe O (6.0 0 ) molecules/mol =.9 0 molecules Fe O Therefore, the number of Fe O molecules needed is.9 0 molecules. (a) The mol ratio of Al:Al O in this reaction is always :. Dividing both the.6 mol of Al and.8 mol of Al O by.8 will give the lowest ratio of :. Your answer is consistent. (b) Likewise, the mol ratio of Al:Fe O in this reaction is always :. Dividing both the mol of Al and mol of Fe O by will give the lowest ratio of :. Your answer is consistent.. Problem The thermal decomposition of ammonium dichromate is an impressive reaction. When heated with a Bunsen burner or propane torch, the orange crystals of ammonium dichromate slowly decompose to green chromium(iii) oxide in a volcano-like display. Colourless nitrogen gas and water vapour are also off. (NH ) Cr O 7(s) Cr O (s) + N (g) + H O (g) (a) Calculate the number of formula units of Cr O that is produced from the decomposition of 0.0 g of (NH ) Cr O 7. (b) In a different reaction, 6.9 g of N is produced when a sample of (NH ) Cr O 7 is decomposed. How many water molecules are also produced in this reaction? (c) How many formula units of (NH ) Cr O 7 are needed to produce.5 g of H O? (a) You need to find the number of molecules of Cr O in the reaction. (b) You need to find the number of water molecules produced in the reaction. (c) You need to find the number of molecules of ammonium dichromate reacted. (a) The mass of ammonium dichromate is 0.0 g. (b) The mass of N is 6.9 g. (c) The mass of water is.5 g. The balanced equation is. Avogadro s number is molecules/mol. (a) Apply the following steps: Step Convert the mass (m) of ammonium dichromate to the number of moles (n), using its (M). Use the formula n = m / M Step Determine the of (NH ) Cr O 7 to Cr O from the balanced equation. Using the result in Step, solve for the number of moles of Cr O. Step Multiply the number of moles of Cr O to the Avogadro number to obtain the number of formula units in the reaction. (b) Repeat the steps in (a), this time working with the mass of N to obtain its number of moles, and then using the of N to H O to obtain the number of moles of water. Multiply this value by the Avogadro number to establish the number of molecules of water produced. (c) Repeat the steps in (a), this time working with the mass of water to obtain its number of moles, and then using the of (NH ) Cr O 7 to H O to obtain the number of moles of (NH ) Cr O 7. Multiply this value by the Avogadro number to establish the number of formula units of (NH ) Cr O 7 needed for the full reaction to proceed. 8

21 (a) (NH ) Cr O 7(s) Cr O (s) + N (g) + H O (g) 5. g/mol 5 g/mol 8.0 g/mol 8.0 g/mol 0.0 g N 0.0 g (NH ) Cr O 5 g/mol mol (NH ) CrO7 ).097 mol (NH) CrO7l mol CrO n mol CrO 7 mol CrO mol (NH ) CrO7l - Step 7 n = =. mol (NH CrO7 0 Step = n = mol (NH ) Cr O n = mol CrO Step N = mol Cr O (6.0 0 ) formula units/mol =.9 0 formula units Cr O Therefore, the number of Cr O molecules produced from the decomposition reaction is.9 0. (b) (NH ) Cr O 7(s) Cr O (s) + N (g) + H O (g) 5. g/mol 5 g/mol 8.0 g/mol 8.0 g/mol 6.9 g N Step n = 6.9 gn 8.0 g/mol = 0.60 mol N Step moln molh O = 0.60 mol N n mol H O n = 0.60 mol N molh O moln =. mol H O Step N =. mol H O (6.0 0 ) molecules/mol =.5 0 molecules H O Therefore, the number of water molecules produced is.5 0. (c) (NH ) Cr O 7(s) Cr O (s) + N (g) + H O (g) 5. g/mol 5 g/mol 8.0 g/mol 8.0 g/mol N.5 g Step n =.5 gh O 8.0 g/mol Step mol(nh ) Cr O 7 molh O = mol H O = n mol (NH ) Cr O mol H O n = mol H O mol(nh ) Cr O 7 = 0.00 mol (NH molh O ) Cr O 7 Step N = 0.0 mol (NH ) Cr O 7 (6.0 0 ) formula units/mol =. 0 formula units (NH ) Cr O 7 Therefore, the number of (NH ) Cr O 7 formula units needed in the reaction is. 0. 8

22 (a) The mol ratio of (NH ) Cr O 7 to Cr O in this reaction is always :. Dividing both the mol of (NH ) Cr O 7 and mol of Cr O by will give the lowest ratio of :. Your answer is consistent. (b) The mol ratio of N :H O in this reaction is always :. Dividing both the 0.00 mol of (NH ) Cr O 7 and mol of H O by 0.00 will give the lowest ratio of :. Your answer is consistent. (c) Likewise, the mol ratio of (NH ) Cr O 7 :H O in this reaction is always :. Dividing both the 0.60 mol of N and. mol of H O by 0.60 will give the lowest ratio of :. Your answer is consistent.. Problem Ammonia gas reacts with oxygen to produce water and nitrogen oxide. This reaction can be catalyzed, or sped up, by Cr O, produced in the reaction in problem. NH (g) + 5O (g) NO (g) + 6H O (l) (a) How many molecules of oxygen are is required to react with.0 g of ammonia? (b) What volume of nitrogen oxide at STP is expected from the reaction of molecules of oxygen with sufficient ammonia? (a) You need to find the number of molecules of oxygen required for the reaction. (b) You need to find the volume of NO produced at STP from the number of oxygen molecules. (a) The mass of the ammonia is.0 g. (b) The number of oxygen molecules is The balanced equation is. Avogadro s number = molecules/mol. (a) Apply the following steps: Step Convert the mass (m) of ammonia to the number of moles (n), using its (M). Use the formula n = m / M Step Determine the of NH to O from the balanced equation. Using the result in Step, solve for the number of moles of O. Step Multiply the number of moles of O by the Avogadro number to obtain the number of molecules in the reaction. (b) Apply the following steps: Step Divide the number of molecules of O by the Avogadro number to obtain the number of moles in the reaction. Step Determine the of O to NO from the balanced equation. Using the result in Step, equate and solve for the number of moles of NO. Step Multiply the number of moles of NO by the molar volume. L mol, to obtain its volume at STP. (a) NH (g) + 5O (g) NO (g) 6H O (l) 7.0 g/mol.0 g 5.00 g/mol N 0.0 g/mol g/mol 85

23 Step.0 g NH n = =.00 mol NH 7. 0 Step mol NH = g/mol.00 mol NH 5 mol O n mol O 5 mol O mol NH n = 00. mol NH = 50. mol O Step N =.50 mol O (6.0 0 ) molecules/mol =.50 0 molecules O Therefore, the number of oxygen molecules required for the reaction is (b) NH (g) + 5O (g) NO (g) 6H O (l) g/mol.00 g/mol 0.0 g/mol 8.0 g/mol molecules m Step Step ( ) molecules O n = =.9 mol O ( ) molecules/mol 5 mol O.9 mol O = mol NO n mol Cr O mol NO n = 9. mol O = 9. mol NO 5 mol O Step m =.9 mol NO. L/mol = 66 L Therefore, the mass of nitrogen oxide expected from the reaction is 57.7 g. (a) The mol ratio of NH to O in this reaction is always :5, which is :.5 in its lowest ratio. Dividing both the.00 mol of NH and.50 mol of O by.00 will give the lowest ratio of :.5. Your answer is consistent. (b) The mol ratio of O :NO in this reaction is always 5:, which is.5: in its lowest ratio. Dividing both the.9 mol of O and.9 mol of NO by.9 will give the lowest ratio of.5:. Your answer is consistent. Solutions for Practice Problems Student Textbook page. Problem The following balanced chemical equation shows the reaction of aluminum with copper(ii) chloride. If 0.5 g of aluminum reacts with 0.5 g of copper(ii) chloride, determine the limiting reactant. Al (s) + CuCl (aq) Cu (s) + AlCl (aq) You need to find the limiting reactant in this reaction. Reactant Al = 0.5 g Reactant CuCl = 0.5 g Products are Cu and AlCl The equation is balanced. Convert the masses into moles (n). From the balanced equation, use the mole ratio of the reactants and products to determine how much either of the products is produced from the calculated number of moles of each reactant. In this case, we will 86

24 choose AlCl. The reactant giving the smallest amount of AlCl (in moles) is the limiting reactant. Al (s) + CuCl (aq) Cu (s) + AlCl (aq) 6.98 g.5 g 6.55 g. g 0.5 g 0.5 g n Number of moles of Al = = 9. 0 Number of moles of CuCl = = g/mol Therefore, the amount of AlCl that should be produced based on: 9. 0 Al Æ n = = 9. 0 mol of AlCl 8. 0 CuCl Æ n = = 5. 0 mol of AlCl CuCl will produce less AlCl than Al, therefore, it is the limiting reactant. The same result should be obtained using Cu as the basis for your calculations instead of AlCl. The of Al:Cu is :; the mol ratio of CuCl :Cu is : mol of Al mol Cu For Al: n of Cu = = 0. mol mol Al 8. 0 mol of CuCl mol CuCl For CuCl : n of Cu = = mol 05. g Al g/mol 0.5 g CuCl mol Cu Again, the CuCl produced less amount of the copper product, making it the limiting reactant. Your result is reasonable.. Hydrogen fluoride, HF, is a highly toxic gas. It is produced by the double displacement reaction of calcium fluoride, CaF, with concentrated sulfuric acid, H SO. CaF (s) + H SO (l) HF (g) + CaSO (s) Determine the limiting reactant when 0.0 g of CaF reacts with 5.5 g of H SO. You need to find the limiting reactant in this reaction. Reactant CaF = 0.0 g Reactant H SO = 5.5 g Products are HF and CaSO The equation is balanced. Convert the masses into moles (n). From the balanced equation, use the mole ratio of the reactants and products to determine how much either of the products is produced from the calculated number of moles of each reactant. In this case, we will choose HF. The reactant giving the smallest amount of HF (in moles) is the limiting reactant. CaF (s) + H SO (aq) HF (g) + + CaSO (s) g g 0.0 g g 0.0 g 5.5 g n 87

25 Number of moles of CaF = = 0. 8 mol Number of moles of HSO = = mol g/mol Therefore, the amount of HF that should be produced based on: 0. 8( ) CaF Æ n = = mol 0. 58( ) HSO Æ n = = 0. 6 mol CaF will produce less HF than H SO, therefore, it is the limiting reactant. The same result should be obtained using CaSO as the basis for your calculations instead of HF. The of CaF :CaSO is :; the mol ratio of H SO :CaSO is : mol of CaF mol CaSO For CaF : n of CaSO = = 0. 8 mol mol CaF 0.6mol of CaSO mol CaSO For H SO : n of CaSO = = 0. 6 mol 0.0 g CaF g/mol 5.5 g H SO mol H SO Again, the CaF produced less amount of the calcium sulfate product, making it the limiting reactant. Your result is reasonable. 5. Acrylic, a common synthetic fibre, is formed from acrylonitrile C H N. Acrylonitrile can be prepared by the reaction of propylene, C H 6, with nitric oxide, NO. C H 6(g) + 6NO (g) C H N (g) + 6H O (g) + N (g) What is the limiting reactant when 6 g of C H 6 reacts with 75 g of NO? You need to find the limiting reactant in this reaction. Reactant C H 6 = 6 g Reactant NO = 75 g Products are C H N, H O, and N The equation is balanced. Convert the masses into moles (n). From the balanced equation, use the mole ratio of the reactants and products to determine how much either of the products is produced from the calculated number of moles of each reactant. In this case, we will choose N. The reactant giving the smallest amount of N (in moles) is the limiting reactant. C H 6(g) + 6NO (g) C H N (g) + 6H O (g) + N (g) g/mol 0.0 g/mol 8.0 g/mol 6 g 75 g n Number of moles of C H 6 = 6 gc H 6 =.99 mol.09 g/mol Number of moles of NO = L NO = 5.85 mol NO. L / mol Therefore, the amount of N that should be produced based on: ( 99. )( ) CH 6 Æ n = = mol ( 58. )( ) HSO Æ n = = mol 6 88

26 Therefore, C H 6 is the limiting reactant. The same result should be obtained using either acrylonitrile or water as the basis for your calculations instead of N. For example, the of C H 6 :H O is :6; the mol ratio of NO:H O is 6:6. For C H 6 : n of H O = =.8 mol For NO: n of H O =.99 mol C H 6 6molH O molc H mol NO 6molH O 6molNO = 5.8 mol Again, the C H 6 produced less amount of the water product, making it the limiting reactant. Your result is reasonable. 6. Problem.76 g of zinc reacts with hydrogen ions as shown in the equation below. Which reactant is present in excess? Zn (s) + H + (aq) Æ Zn + (aq) + H (g) You need to find the reactant in excess in the reaction. Reactant Zn =.76 g Reactant HCl = molecules. You know Avogadro s number = molecules/mol. Step Write the balanced equation for this reaction. Step Determine the number of moles (n) of each reactant. For Zn, divide the mass by its. For HCl, divide the number of molecules by the Avogadro number. Step From the balanced equation, use the of the reactants and products to determine how much either of the products is produced from the calculated number of moles of each reactant. In this case, we will choose H. The reactant giving the largest amount of H (in moles) is the reactant in excess. Zn + HCl ZnCl + H 65.9 g/mol 6.6 g/mol 6.9 g/mol.0 g/mol.76 g (8.9)(0 ) molecules n Step The balanced equation is Zn (s) + H + aq Æ Zn + (aq) + H (g) Step Number of moles of Zn =.76 g Zn = mol molecules HCl molecules/mol Number of moles of HCl = =.8 mol Step Therefore, the amount of H that should be produced based on: ( ) Zn Æ n = = mol HCl n = (0.7)() = 0.7 mol Therefore, HCl is present in excess g/mol 89

27 The same result should be obtained using zinc chloride as the basis for your calculations instead of hydrogen. For example, the of Zn:ZnCl is :; the mol ratio of HCl:ZnCl is : mol of Zn mol ZnCl For Zn: n of ZnCl = = mol mol Zn.8 mol of HCl mol ZnCl For HCl: n of ZnCl = = mol Again, the HCl produced more ZnCl product, making it the reactant in excess. Your result is reasonable. Solutions for Practice Problems Student Textbook page 5 mol HCl 7. Problem Chlorine dioxide, ClO, is a reactive oxidizing agent. It is used to purify water. 6ClO (g) + H O (l) 5HClO (aq) + HCl (aq) (a) If 7.00 g of ClO is mixed with 9.00 g of water, what is the limiting reactant? (b) What mass of HClO expected in part (a)? (c) How many molecules of HCl are expected in part (a)? (a) You need to find the limiting reactant in the reaction. (b) You need to find the mass of the HClO produced in the reaction. (c) You need to find the number of molecules of HCl produced in the reaction. Reactant ClO = 7.00 g Reactant H O = 9.00 g Products are HClO and HCl The balanced equation is. You know Avogadro s number = molecules/mol. (a) Convert the masses into moles (n). From the balanced equation, use the of the reactants and products to determine how much either of the products is produced from the calculated number of moles of each reactant. In this case, we will choose HClO. The reactant giving the smallest amount of HClO (in moles) is the limiting reactant. (b) Multiply the mole amount of HClO, obtained from the limiting reactant in (a), by the of HClO to obtain its mass in grams. (c) Repeat step (a) to solve for the number of moles of HCl produced by the limiting reactant. Multiply this result by the Avogadro number to obtain the number of molecules of HCl produced. 6ClO (g) + H O (l) 5HClO (aq) + HCl (aq) g/mol 8.0 g/mol 8.6 g/mol 6.6 g/mol 7.00 g 9.00 g (a) Number of moles of ClO Number of moles of HO = =.05 mol 7.00 g ClO g/mol = =.05 mol 9.00 g Zn 8. 0 g/mol 90

28 Therefore, the amount of HClO that should be produced based on: ClO Æ n = = mol HO Æ n = =. 757 mol Therefore, ClO is the limiting reactant. (b) Number of moles of HClO produced = mol (from (a)) Therefore, m = mol HClO 8.6 g/mol = 7.97 g ClO (c) Using the number of moles of ClO (the limiting reactant):. 05 Number of moles of HCl produced = mol 6 Therefore, N = 0.75 mol HCl (6.0 0 ) molecules/mol = molecules HCl The of the reactants ClO :H O was 6: which is : in its lowest ratio. The calculated from the masses of reactants was mol ClO :.757 mol H O, which is : in its lowest ratio (dividing both by ). Clearly, there was not enough ClO available to satisfy the ratio needed by the balanced equation, therefore, it is the limiting reactant in this case. The results are reasonable. 8. Problem Hydrazine, N H, reacts exothermically with hydrogen peroxide, H O. N H (l) + 7H O (aq) HNO (g) + 8H O (g) (a) 0 g of N H reacts with an equal mass of H O. Which is the limiting reactant? (b) What mass of HNO is expected? (c) What mass, in grams, of the excess reactant remains at the end of the reaction? (a) You need to find the limiting reactant in the reaction. (b) You need to find the mass of the HNO produced in the reaction. (c) You need to find the mass of excess reactant left over after the reaction. Reactant N H = 0 g Reactant H O = 0 g Products are HNO and H O The balanced equation is. (a) Convert the masses into moles (n). From the balanced equation, use the of the reactants and products to determine how much either of the products is produced from the calculated number of moles of each reactant. In this case, we will choose HNO. The reactant giving the smallest amount of HNO (in moles) is the limiting reactant. (b) Multiply the mole amount of HNO, obtained from the limiting reactant in (a), by the of HNO to obtain its mass in grams. (c) Using the of the reactants N H and H O from the balanced equation and the number of moles of the limiting reactant calculated in (a), equate and solve for the number of moles of excess reactant used in this reaction. Subtract this mole amount from the mole amount calculated in (a) for the identified excess reactant. Multiply the mole difference by the of the excess reactant to obtain the mass left over. 9

29 N H (l) + 7H O (aq) HNO (g) + 68H O (g) g 0 g n (a) Number of moles of N H = 0 gn H =.7 mol.06 g/mol 0 g HO Number of moles of HO = =. 5 mol. 0 g/mol Therefore, amount of HNO that should be produced based on: N H =.7 = 7.8 mol.5 HO = = 0. mol 7 H O is the limiting reactant. (b) Number of moles of HNO produced =.0 mol (from (a)) Therefore, m =.0 mol HNO 6.0 g/mol = 6.6 g HNO (c) Number of moles of N H used =.5 mol 7 = 0.50 mol Therefore, number of moles of N H in excess=.7 mol 0.50 mol=. mol Hence, m =. mol N H.06 g/mol = 0 g The mass of N H left over after the reaction is 0 g. The of the reactants N H to H O from the balanced equation is :7. The calculated from the masses of reactants was.7 mol N H :.5 mol H O, which is.06: in its lowest ratio (dividing both by.5). Clearly, there was not enough H O available to satisfy the :7 ratio needed by the balanced equation, therefore, it is the limiting reactant in this case and the N H was in excess. The results are reasonable. 9. Problem In the textile industry, chlorine is used to bleach fabrics. Any of the toxic chlorine that remains after the bleaching process is destroyed by reacting it with a sodium thiosulfate solution, Na S O (aq). Na S O (aq) + Cl (g) + 5H O (l) NaHSO (aq) + 8HCl (aq) 5 kg of Na S O reacts with 50.0 kg of Cl and 8 kg of water. How many grams of NaHSO are expected to be produced? You need to find the mass of the NaHSO produced in this reaction. Reactant Na S O = 5 kg Reactant Cl = 50.0 kg Reactant H O = 8 kg Products are NaHSO and HCl The balanced equation is. Follow the steps below: Step Determine the identity of the limiting reactant for the amounts of reactants. Convert the masses into moles (n). From the balanced equation, use the of the reactants and products to determine how much either of the products is produced from the calculated number of moles of each reac- 9

30 tant. In this case, we will choose NaHSO. The reactant giving the smallest amount of NaHSO (in moles) is the limiting reactant. Remember to convert the kg quantities to grams before proceeding with the calculations. Step From Step, the number of moles of NaHSO produced by the limiting reactant is the amount that will be produced by this reaction. Multiply this mole amount by the of NaHSO to obtain its mass in grams. Na S O (aq) + Cl (g) + 5H O (l) NaHSO + 8HCl (aq) Mole ratio g/mol 70.9 g/mol 8.0 g/mol 0.07 g/mol 6.6 g/mol 5 kg 50.0 kg 8 kg n Step Number of moles of Number of moles of Number of moles of HO = =. 0 mol Therefore, the mole amount of NaHSO that should be produced based on: 85 mol Na SO Æ n = = 7. 0 mol 708 mol Cl Æ n = = 5 mol. 0 mol HO Æ n = = Na S O = = 85 mol ( ) g Cl g/mol 5 ( 8. 0 ) g H O g/mol Cl = = mol (.5 0 ) g Na S O 58. g/mol The limiting reactant is Cl. Step From Step, 5 mol of NaHSO will be produced. Therefore, m = 5 mol NaHSO 0.07 g/mol =.5 0 g. The mass of NaHSO produced is.5 0 g. The of the reactants Na S O to Cl to H O from the balanced equation is ::5. The calculated from the masses of reactants was :. 0, which is.::8.7 in its lowest ratio (dividing all three by 708). Clearly, there is not enough Cl available to satisfy the : ratio needed between it and Na S O, therefore, Cl is the limiting reactant in this case. The results are reasonable. 0. Problem Manganese(III) fluoride can be formed by the reaction of manganese(ii) iodide with fluorine. MnI (s) + F (g) MnF (s) + IF 5(l) (a). g of MnI reacts with 5.0 g of F. What mass of MnF is expected? (b) How many molecules of IF 5 are produced in part (a)? (c) What reactant is in excess? How much of it remains at the end of the reaction? (a) You need to find the mass of the MnF produced in this reaction. (b) You need to find the number of IF 5 molecules produced in this reaction. (c) You have to find the mass of excess reactant left over after the reaction. Reactant MnI =. g Reactant F = 5.0 g Products are MnF and IF 5 5 9

31 The balanced equation is. The Avogadro constant = molecules/mol (a) Determine the identity of the limiting reactant for the amounts of reactants. Convert the masses into moles (n). From the balanced equation, use the mole ratio of the reactants and products to determine how much MnF is produced from the calculated number of moles of each reactant. The reactant giving the smallest amount of MnF (in moles) is the limiting reactant. Take this smaller mole amount of MnF and multiply it by its to obtain the mass in grams. (b) Repeat (a) to establish the of the limiting reactant to IF 5 from the balanced equation, and equate and solve for the number of moles of IF 5. Multiply this mole amount by the Avogadro constant to establish the number of molecules of IF 5 produced. (c) From (a), the excess reactant can be identified. Using the of MnI and F from the balanced equation and the number of moles of limiting reactant used in the reaction, equate and solve for the mole amount of the excess reactant. Subtract this value from the mole amount at the start of the reaction to obtain the mole amount in left over. MnI (s) + F (g) MnF (g) + IF 5(l) 08.7 g/mol 8.00 g/mol.9 g/mol.90 g/mol. g.7 L m - (a) Number of moles of MnI = = mol 08.7 g/mol Number of moles of F =. 7 L = mol. L / mol Therefore, the number of moles of MnF that should be produced based on: mol - MnI Æ n = = mol F n = mol. g MnI = 0.0 mol The limiting reactant is MnI. Using the mole amount produced by the limiting reactant: m = mol MnF.9 g/mol = 0.6 g MnF Therefore, 0.6 g of MnF will be produced mol - (b) Number of moles of IF 5 produced = mol N = mol IF 5 (6.0 0 ) molecules/mol =.79 0 molecules Therefore,.8 0 molecules of IF 5 will be produced. (c) From (a), the reactant in excess is F mol Number of moles of F used = mol Therefore, number of moles of F in excess=0.658 mol-0.06 mol=0.6 mol. Therefore mass of F in excess = 0.6 mol 8.0 g/mol =.0 g. The of the reactants MnI to F from the balanced equation is :, which is in its lowest ratio. The calculated from the masses of reactants was 0.00 mol MnI :0.658 mol F, which is :6 in its lowest ratio (dividing both by ). Clearly, there is not enough MnI available to - 9

32 react with the vast amount of F present at the start of the reaction. Therefore, MnI is the limiting reactant in this case. The results are reasonable. Solutions for Practice Problems Student Textbook page 9. Problem 0.0 g of bromic acid, HBrO, are reacted with excess HBr. HBrO (aq) + 5HBr (aq) H O (l) + Br (aq) (a) What is the theoretical yield of Br for this reaction? (b) If 7. g of Br are produced, what is the percentage yield of Br? (a) You need to find the theoretical yield of Br in the reaction. (b) You need to find the actual percentage yield of Br from the mass produced. (a) Mass of reactant bromic acid = 0.0 g. (b) Actual yield of Br = 7. g. The balanced equation is. (a) Divide the mass of bromic acid by its to obtain the mole amount. Use this value and the of HBrO to Br from the balanced equation to equate and solve for the number of moles of Br produced. Multiply this value by the of Br to obtain the theoretical yield in grams. (b) Divide the actual yield by the theoretical yield calculated in (a) and multiply by 00%. HBrO (aq) + 5HBr (aq) H O (l) + Br (aq) g/mol 80.9 g/mol 8.0 g/mol 59.8 g/mol 0.0 g m = 0.55 mol (a) Number of moles of HBrO = 0.0 ghbro 8.9 g/mol Therefore, the number of moles of Br produced = Hence, m = 0.65 Br 59.8 g/mol = 7. g The theoretical yield of Br is 7. g. (b) Percentage yield of Br = 7. g 00% = 6. 7 mol 7. g Use whole numbers for a quick inspection: 7 7 consistent with the percentage result mol = 0.65 mol = This decimal result is. Problem Barium sulfate forms as a precipitate in the following reaction: Ba(NO ) (aq) + NaSO (aq) BaSO (s) + NaNO (aq) When 5.0 g of Ba(NO ) are reacted with excess NaSO, 9.8 g of BaSO is recovered by the chemist. (a) Calculate the theoretical yield of BaSO. (b) Calculate the percentage yield of BaSO. 95

33 (a) You need to find the theoretical yield of BaSO in the reaction. (b) You need to find the actual percentage yield of BaSO from the mass recovered. (a) Mass of reactant Ba(NO ) = 5.0 g. (b) Actual yield of BaSO = 9.8 g. The balanced equation is. (a) Divide the mass of Ba(NO ) by its to obtain the mole amount. Use this value and the of Ba(NO ) to BaSO from the balanced equation to equate and solve for the number of moles of BaSO produced. Multiply this value by the of BaSO to obtain the theoretical yield in grams. (b) Divide the actual yield by the theoretical yield calculated in (a) and multiply by 00%. Ba(NO ) (aq) + Na SO (aq) BaSO (s) + NaNO (aq) 6.5 g/mol 5.0 g.05 g/mol. g/mol m 85 g/mol (a) Number of moles of Ba(NO ) = 5.0 gba(no ) 6.5 g/mol Therefore, the number of moles of BaSO produced = Hence, m = 0. mol BaSO. g/mol =. g The theoretical yield of BaSO is. g. 9.8 g (b) Percentage yield of BaSO = 00% = 95. %. g Use whole numbers for a quick inspection: 0 consistent with the percentage result. = 0. mol 0. mol = This decimal result is = 0. mol. Problem Yeasts can act on a sugar, such as glucose, C 6 H O 6, to produce ethyl alcohol, C H 5 OH, and carbon dioxide. C 6 H O 6 C H 5 OH + CO If g of ethyl alcohol are recovered after.6 kg of glucose react, what is the percentage yield of the reaction? You need to find the percentage yield of this reaction. Reactant glucose =.6 kg. Actual yield of ethyl alcohol = g. The balanced equation is. First find the theoretical yield: Divide the mass of glucose by its to obtain the mole amount. Use this value and the of glucose to ethyl alcohol from the balanced equation to equate and solve for the number of moles of ethyl alcohol produced. Multiply this value by the of ethyl alcohol to obtain the theoretical yield in grams. Finally, divide the actual yield by the theoretical yield and multiply by 00%. 96

34 C 6 H O 6 C H 5 OH + CO 80.8 g/mol 6.08 g/mol.0 g/mol.6 kg m Number of moles of C 6 H O 6 Number of moles of C H 5 OH produced = 8. mol Therefore, m = 8. mol C H 5 OH 6.08 g/mol = 8 C H 5 OH produced g Finally, the percentage yield of C H 5 OH 00 % = 6. 7 % 8 g Use whole numbers for a quick inspection: 00 = 0.5. This decimal result is 800 consistent with the percentage result. Solutions for Practice Problems Student Textbook page (.6 0 ) g C S O 80.8 g/mol 6 6 = 905. mol 905. mol. Problem The following reaction proceeds with a 70% yield. C 6 H 6(l) + HNO (aq) C 6 H 5 NO (l) + H O (l) Calculate the mass of C 6 H 5 NO expected if.8 g of C 6 H 6 expected if.8 g of C 6 H 6 reacts with excess HNO. You have to find the actual yield of C 6 H 5 NO in this reaction. Reactant C 6 H 6 =.8 g. Percentage yield of C 6 H 5 NO is 70%. The balanced equation is. First find the theoretical yield: Divide the mass of C 6 H 6 by its to obtain the mole amount. Use this value and the of C 6 H 6 to C 6 H 5 NO from the balanced equation to equate and solve for the number of moles of C 6 H 5 NO produced. Multiply this value by the of C 6 H 5 NO to obtain the theoretical yield in grams. Finally, multiply the theoretical yield by the percentage yield (converted to a decimal) to obtain the actual yield in grams. C 6 H 6(l) + HNO (aq) C 6 H 5 NO (l) + H O (l) 78. g/mol 6.0 g/mol. g/mol 8.0 g/mol.8 g m.8 g Number of moles of C 6 H 6 = = 0. 6 mol 76. g/mol Number of moles of C 6 H 5 NO produced = 0.6 mol = 0.6 mol Therefore, m = 0.6 mol C 6 H 5 NO. g/mol = 0. g C 6 H 5 NO Finally, the actual yield = 0. g C 6 H 5 NO =. g 97

35 Use whole numbers for a quick inspection: =.0. The results match. 5. Problem The reaction of toluene, C 7 H 8, with potassium permanganate, KMnO, gives less than a 00% yield. C 7 H 8(l) + KMnO (aq) KC 7 H 5 O (aq) + MNO (s) + KOH (aq) + H O (l) (a) 8.60 g of C 7 H 8 is reacted with excess KMnO. What is the theoretical yield, in grams, of KC 7 H 5 O? (b) If the percentage yield is 70.0%, what mass of KC 7 H 5 O can be expected? (c) What mass of C 7 H 8 is needed to produce. g of KC 7 H 5 O, assuming a yield of 60%? (a) You need to find the theoretical yield of KC 7 H 5 O. (b) You need to find the mass of KC 7 H 5 O assuming the percentage yield. (c) You need to find the mass of C 7 H 8 that yields the give percentage yield of product. (a) Reactant C 7 H 8 = 8.6 g (b) Percentage yield of product KC 7 H 5 O = 70% (c) Product KC 7 H 5 O =. g at a percentage yield of 60% The balanced equation is. (a) Divide the mass of C 7 H 8 by its to obtain the mole amount. Use this value and the of C 7 H 8 to KC 7 H 5 O from the balanced equation to equate and solve for the number of moles of KC 7 H 5 O produced. Multiply this value by the of KC 7 H 5 O to obtain the theoretical yield in grams. (b) Multiply the theoretical yield in (a) by the percentage yield (converted to a decimal) to obtain the actual yield in grams. (c) Work backward with the values. Divide the theoretical yield by the percentage yield (converted to a decimal) to obtain the actual yield in grams. Dive this mass by the of KC 7 H 5 O to obtain its number of moles. Use this value and the of C 7 H 8 to KC 7 H 5 O from the balanced equation to equate and solve for the number of moles of C 7 H 8. Multiply this value by the of C 7 H 8 to obtain the mass. C 7 H 8(l) + KMnO (aq) KC 7 H 5 O (aq) + MnO (s) + KOH (aq) + H O (l) 9.5 g/mol 8.60 g 58.0 g/mol 60. g/mol m 86.9 g/mol 56. g/mol 8.0 g/mol 8.60 g C H 7 8 (a) Number of moles of C 7 H 8 = = mol 9.5 g/mol 0.9 g mol Number of moles of KC 7 H 5 O produced = = mol Therefore, m = 0.09 mol KC 7 H 5 O 60. g/mol =.9 g The theoretical yield of KC 7 H 5 O is.9 g. (b) Expected yield = actual yield =.9 g KC 7 H 5 O 0.07 = 0. g KC 7 H 5 O (c) Actual yield of KC 7 H 5 O =. g KC 7H 5 O =. g 0.6 Number of moles of KC 7 H 5 O =. gkc 7H 5 O = 0.9 mol 60. g/mol 98

36 Number of moles C 7 H 8 required = 0.9 mol = 0.9 mol m = 0.9 mol C 7 H g/mol =.8 g The mass of C 7 H 8 required is.8 g. 0. g (a) and (b). The number of moles of actual KC 7 H 5 O = = mol. 60. g/mol The theoretical number of moles calculated = 0.9 mol Actual mole / theoretical mol = = = 69. 6%. This is consistent with the 70% percentage yield. (c) Likewise, the theoretical number of moles of. g. g 60. g/mol KC 7 H 5 O = = mol. The actual number of moles calculated = 0.9 mol. Actual mole / theoretical mol = = 0.60 = 60.%. This is consistent with 0.9 the 60% percentage yield. 6. Problem Marble is made primarily of calcium carbonate. When calcium carbonate reacts with hydrogen chloride, it forms calcium chloride, carbon dioxide, and water. If this reaction occurs with 8.5% yield, what mass of carbon dioxide will be collected if 5.7 g of CaCO is added to sufficient hydrogen chloride? You need to find the actual yield of CO from the reaction. Reactant CaCO = 5.7 g. Percentage yield of CO = 8.5%. The reactants and products of the reaction are. Step Write the full balanced equation for this reaction. Step Determine the theoretical yield of CO. Divide the mass of CaCO by its to obtain the mole amount. Use this value and the of CaCO to CO from the balanced equation to equate and solve for the number of moles of CO produced. Multiply this value by the of CO to obtain the theoretical yield in grams. Step Multiply the theoretical yield by the percentage yield (converted to a decimal) to obtain the actual yield in grams. CaCO (s) + HCl (aq) CaCl (s) + CO (g) + H O (l) g/mol 6.6 g/mol 0.98 g/mol.0 g/mol 8.0 g/mol 5.7 g m Step Balanced equation: CaCO (s) + HCl (aq) CaCl (s) + CO (g) + H O (l) Step Number of moles of CaCO = Number of moles of CO produced = Theoretical yield of CO = 0.57 mol. L/mol =.5 g Step Actual yield =.5 L 0.85 = g CaCO = mol CaCO g/mol 0.57 g mol = mol CO 99

37 5.6 g CO = mol. = 0.8 of actual CO produced..0 g/mol 0. 8.Actual mole / theoretical mol = = 8. 5%. The answer is reasonable Problem Mercury, in its elemental form or in a chemical compound is highly toxic. Water-soluble mercury compounds, such as mercury(ii) nitrate, can be removed from industrial wastewater by adding sodium sulfide to the water, which forms a precipitate of mercury(ii) sulfide, which can then be filtered out. Hg(NO ) (aq) + Na S (aq) HgS (s) + NaNO (aq) If.5 0 formula units of Hg(NO ) are reacted with excess Na S, what mass of Na S, what mass of HgS can be expected if this process occurs with 97.0% yield? You need to find the actual yield of HgS produced. Reactant Hg(NO ) =.5 0 formula units. The percentage yield of product HgS = 97.0%. The balanced equation is. You know Avogadro s number = formula units/mol. Step Determine the theoretical yield of HgS. Divide the number of formula units of Hg(NO ) by Avogadro s number to obtain the mole amount. Use this value and the of Hg(NO ) to HgS from the balanced equation to equate and solve for the number of moles of HgS produced. Multiply this value by the of HgS to obtain the theoretical yield in grams. Step Multiply the theoretical yield by the percentage yield (converted to a decimal) to obtain the actual yield in grams. Hg(NO ) (aq) + Na S (aq) HgS (s) + NaNO (aq).6 g/mol g/mol.66 g/mol 85 g/mol.5 0 formula units m (.5 0 Step Number of moles of Hg(NO ) = ) formula units (6.0 0 ) formula units/mol = 0.57 mol Number of moles of HgS produced = = 0.57 mol 0.57 mol Theoretical yield, m = 0.57 mol HgS.66 g/mol = g HgS. Step Actual yield = g HgS 0.07 = 9 g HgS 9 g CO = mol of actual HgS produced..66 g/mol Actual mole / theoretical mol = = = 96. 7%. This is equal to the value in the question. 00

38 Solutions for Practice Problems Student Textbook pages Problem An impure sample of silver nitrate, AgNO, has a mass of 0.0 g. It is dissolved in water and then treated with excess hydrogen chloride, HCl (aq). This results in the formation of a precipitate of silver chloride, AgCl. AgNO (aq) + HCl (aq) AgCl (s) + HNO (aq) The silver chloride is filtered, and any remaining hydrogen chloride is washed away. Then the silver chloride is dried. If the mass of the dry silver chloride is measured to be 0. g, what mass of silver nitrate was contained in the original (impure) sample? You need to find the total mass of the original (impure) silver nitrate. Reactant silver nitrate = 0.0 g. Actual yield of the silver chloride = 0. g. The balanced equation is. You can assume the reaction proceeds to completion under the excess HCl. Divide the actual yield of silver chloride by its to obtain the number of moles. Use this value and the of AgNO and AgCl from the balanced equation to equate and solve for the number of moles of AgNO. Multiply this value by the of silver nitrate to obtain the mass. AgNO (aq) + HCl (aq) AgCl (s) + HNO (aq) g/mol 6.6 g/mol. g/mol 6.0 g/mol m 0. g 0. g CaCO Number of moles of of AgCl produced = = mol. g/mol g mol Number of moles of AgNO required = = mol Therefore, m = mol AgNO g/mol = 0.5 g.. Therefore, 0.5 g of silver nitrate in the original sample. Note that the mass of the original sample was not needed for the calculation. Work backwards. The mass of silver nitrate calculated divided by the number of moles 0. 5 should give the. = 69. g/mol. This is close to the value in the calculations. Your answer is reasonable. 9. Problem Copper metal is mined as one of the several copper-containing ores. One of these ores contains copper in the form of malachite. Malachite exists as a double salt, Cu(OH) CuCO. It can be thermally decomposed as 00 C to yield copper(ii) oxide, carbon dioxide gas, and water vapour. Cu(OH) CuCO (s) CuO (s) + CO (g) + H O (g) (a) kg of malachite ore, containing 5.0% malachite, Cu(OH) CuCO, is thermally decomposed. Calculate the mass of copper(ii) oxide that is formed. Assume 00% reaction. 0

39 (b) Suppose that the reaction has a 78.0% yield, due to incomplete decomposition. How many grams of CuO would be produced? (a) You need to find the theoretical yield of CuO formed in the reaction. (b) You need to find the actual yield of CuO at the percentage yield. (a) Malachite ore = kg; malachite composition in ore = 5.0%. Reaction proceeds to 00%. (b) Percentage yield of CuO is 78.0%. (a) Multiply the mass of malachite ore by the percentage (expressed as a decimal) to get the mass of the malachite. Convert the kg quantity to gram. Divide this mass in grams by the of malachite to obtain the number of moles. Use this value and the of Cu(OH) CuCO and CuO from the balanced equation to equate and solve for the number of moles of CuO. Multiply this value by the of CuO to obtain the theoretical yield in grams. (b) Multiply the theoretical yield obtained in (a) by the percentage yield (expressed as a decimal) to get the actual yield in grams. Cu(OH) CuCO (s) CuO (s) + CO (g) + H O (g). g/mol g/mol.0 g/mol 8.0 g/mol kg (a) Mass of Cu(OH) CuCO in malachite ore = g 0.05 = 60 g Number of moles of Cu(OH) CuCO = 60 gcu(oh) CuCO =.8 mol. g/mol Number of moles of CuO produced = =.6 mol.8 mol Mass CuO produced =.6 mol g/mol = 88 g The theoretical yield of CuO is 88 g. (b) Actual yield = 88 g CuO 0.78 = 7 g. The actual yield in mole amount is 7 g = 85. mol. Actual mole theoretical mol = = = 78.% g/mol This closely matches the percentage yield of 78% in the question. The result is reasonable. 0. Problem Ethylene oxide, C H O, is a multi-purpose industrial chemical used, among other things, as a rocket propellant. It can be prepared by reacting ethylene bromohydrin, C H 5 OBr, with sodium hydroxide. C H 5 OBr + NaOH C H O + NaBr + H O If this reaction proceeds with an 89% yield, what mass of C H O can be obtained when.6 0 molecules of C H 5 OBr react with excess sodium hydroxide? You need to find the actual yield of the C H O produced. 0

40 Reactant C H 5 OBr =.6 0 molecules Reactant NaOH is in excess Percentage yield of C H O = 89% You are the balanced equation. You know Avogadro s number = molecules/mol. First determine the theoretical yield. Divide the number of molecules of C H 5 OBr by the Avogadro number to obtain its number of moles. Use this value and the of C H 5 OBr and C H O from the balanced equation to equate and solve for the number of moles of C H O. Multiply this value by the of C H O to obtain the theoretical yield in grams. Multiply the theoretical yield by the percentage yield (expressed as a decimal) to get the actual yield in grams. C H 5 OBr +NaOH C H O + NaBr + H O.97 g/mol 0 g/mol.06 g/mol 0.89 g/mol 8.0 g/mol.6 0 molecules m Number of moles of C H 5 OBr = Number of moles of C H O produced = = 06. mol Hence, m of C H O = mol.06 g/mol = 6. g Actual yield = 6. g C H O Q 0.89 Q=.5 C H O The actual yield in mole amount is = 0.5 mol molecules CH5OBr molecules/mol mol.5 g.06 g/mol Actual mole = = = 88.8%. This closely matches the theoretical mol percentage yield of 89% in the question. The result is reasonable. = mol 0