Quantities in Chemical Reactions

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 6.0 0 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

H (g) + O (g) H O (l) (a) (b) (c) (d) 00 molecules 50 molecules 78 molecules 6.0 0 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 6.0 0 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 6.0 0 molecules of oxygen by 6.0 0 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.806 0 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

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.806 0 molecules.806 0 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 =.806 0 molecules of Cl (d) molecules of Cl molecules of FeCl =.806 0 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.806 0 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.806 0 molecules of Cl and.0 0 molecules of FeCl by.0 0 will give the lowest ratio of.5:. Your answer is consistent. 66

. 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 6.7 0 5 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) 6.0 0. 0 6.0 0. 0 6.7 0 5 N 6.7 0 5 N (a) 5 60. 0. 0 N = ( 67. 0 ) =. 0 formula units of CaCl 5 (b) N = (6.7 0 5. 0 ) = 6.7 0 5 molecules H. 0 O (a) The of HCl to CaCl in this reaction is always :. Dividing both the 6.7 0 5 molecules of HCl and.5 0 5 molecules of water by.5 0 5 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 6.7 0 5 of HCl and 6.7 0 5 molecules of water by 6.7 0 5 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

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

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) 7 6.9 mol n.0 mol n 69

(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 = 6.0 0 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

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

(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 9.5 0 - mol of N to form N O? (b) How many moles of O are required to react with 9.5 0 - 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

moln molo = 9.5 0- mol N n mol O - mol O - mol N n = ( 95. 0 ) mol N = 68. 0 mol O (b) N (g) + O (g) NO (g) (9.5 )(0 - ) mol n moln = 9.5 0- mol N molo n mol O - mol O mol N n = 95. 0 mol N = 0. 87 mol O (a) The mol ratio of N :O in this reaction is always :. Dividing both the 9.5 0 - mol of N and.68 0 - mol of O by.68 0 - 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 9.5 0 - mol of N and 0.9 mol of O by 9.5 0 - 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.5 0 - 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

(a) Xe (g) + F (g) XeF (s).5 0 - mol n (b) molxe =.5 0- mol Xe molf n mol F n = (.5 0 - ) mol Xe molf molxe = 7.08 0- mol F Xe (g) + F (g) XeF 6(s).5 0 - mol n molxe =.5 0- mol Xe molf n mol F n = (.5 0 - ) 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 0.708 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

(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 = 0. 66 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) 6 59.7 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 = 0 66 6 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

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) 55.85 g/mol 6.6 g/mol 6.75 g/mol.0 g/mol.56 g m.56 gfe 55.85 g/mol Step n = = 0.067 mol Fe Step molfe 0.067 mol Fe = molhcl n mol HCl n = 0 067 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 0.067 mol of Fe and 0.7 mol of HCl by 0.067 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

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 0.0509 mol NO 6.0g/mol mol NO 0. 0509 mol NO = mol O n mol O mol O mol NO n = = n = 0. 0509 mol O = 0. 07 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 0.0509 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

8Zn (s) + S 8(s) 8ZnS (s) 8 8 65.9 g/mol 56.56 g/mol 97.6 g/mol.0 g m Step Step.0 g S 0.5 mol S 56.56 g/mol 8 mol S8 0. 5 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.76 0 - 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 0. 00 mol HCl = mol Cl n mol HCl n = = 0.00 mol HCl mol Cl mol HCl n = 0. 00 mol HCl =.8 0 mol Cl Step m =.8 0 mol Cl. L/mol = 7.5 0 L 78

The mol ratio of HCl:Cl in this reaction is always :. Dividing both the 0.00 mol of HCl and.8 0 - 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 0. 07 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

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 5 6 9. 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

The mass of the nitrogen produced is. The balanced equation is. Avogadro s number, N A = 6.0 0 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

(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 = 6.0 0 formula units/mol. (b) The mass of the aluminum is. The balanced equation is. Avogadro s number, N A = 6.0 0 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 55.85 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 55.85 g/mol 0. g N 0. g Al - Step n = =.97 0 mol Al 6. 98 g/mol mol Al -.97 0 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

Step N =.8 0 - 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.8 0 - mol of Al and.8 0 - mol of Fe O by.8 0 - 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 6.0 0 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

(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 0.097 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 = 0.097 mol (NH ) Cr O n = 97. 0 mol CrO Step N =.97 0 - 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 = 0.0805 mol H O = n mol (NH ) Cr O 7 0.0805 mol H O n = 0.0805 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

(a) The mol ratio of (NH ) Cr O 7 to Cr O in this reaction is always :. Dividing both the 0.096 mol of (NH ) Cr O 7 and.97 0 - mol of Cr O by 0.096 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 0.0805 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 8.95 0 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 8.95 0. The balanced equation is. Avogadro s number = 6.0 0 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 6 8.0 g/mol 85

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.50 0. (b) NH (g) + 5O (g) NO (g) 6H O (l) 5 6 7.0 g/mol.00 g/mol 0.0 g/mol 8.0 g/mol 8.95 0 molecules m Step Step ( 895. 0 ) molecules O n = =.9 mol O ( 60. 0 ) 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

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 = = 8. 0.5 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 :. 9. 0 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 = = 0. 008 mol 05. g Al 6. 98 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) 78.08 g 98.09 g 0.0 g 98.09 g 0.0 g 5.5 g n 87

Number of moles of CaF = = 0. 8 mol Number of moles of HSO = = 0. 58 mol 98.09 g/mol Therefore, the amount of HF that should be produced based on: 0. 8( ) CaF Æ n = = 0. 56 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 :. 0. 8 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 78.08 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) 6 6.09 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 = = 0. 78 mol ( 58. )( ) HSO Æ n = = 0. 97 mol 6 88

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 6 5.8 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 8.9 0 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 = 8.9 0 molecules. You know Avogadro s number = 6.0 0 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 = 0.0575 mol 8.9 0 molecules HCl 6.0 0 molecules/mol Number of moles of HCl = =.8 mol Step Therefore, the amount of H that should be produced based on: 0. 0575( ) Zn Æ n = = 0. 0575 mol HCl n = (0.7)() = 0.7 mol Therefore, HCl is present in excess. 65. 9 g/mol 89

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 :. 0.0575 mol of Zn mol ZnCl For Zn: n of ZnCl = = 0. 0575 mol mol Zn.8 mol of HCl mol ZnCl For HCl: n of ZnCl = = 0. 70 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 = 6.0 0 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) 6 5 67.5 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 67. 5 g/mol = =.05 mol 9.00 g Zn 8. 0 g/mol 90

Therefore, the amount of HClO that should be produced based on:. 05 5 ClO Æ n = = 0. 8758 mol 6. 05 5 HO Æ n = =. 757 mol Therefore, ClO is the limiting reactant. (b) Number of moles of HClO produced = 0.8758 mol (from (a)) Therefore, m = 0.8758 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 = 0. 75 mol 6 Therefore, N = 0.75 mol HCl (6.0 0 ) molecules/mol =.055 0 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 0.8758 mol ClO :.757 mol H O, which is : in its lowest ratio (dividing both by 0.8758). 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