# Ch.4 Solutions. Solutions for Practice Problems p Consider the following reaction. 2H 2(g) + O 2(g) 2H 2 O (l)

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1 Solutions for Practice Problems p. 114 Ch.4 Solutions 1. Consider the following reaction. 2H 2(g) + O 2(g) 2H 2 O (l) (a) Write the ratio of H 2 molecules: O 2 molecules: H 2 Omolecules. The ratio is given by the whole number in front of each reactant and product in the balanced equation. The ratio is 2: 1: 2 (b) How many molecules of O 2 are required to react with 100 molecules of H 2, according to your ratio in part (a) 2 molecules H 2 : 1 molecule O 2, (half),therefore 100 molecules H 2 : 50 molecules O 2 (c) How many molecules of water are formed when 2478 molecules of O 2 react with H 2? 1 molecule O 2 :2 molecules H 2 O, (double), therefore 2478 molecules O 2 : 4956 molecules H 2 O (d) How many molecules of H 2 are required to react completely with 6.02 x molecules of O 2? 2 molecules H 2 : 1 molecule O 2 (double), therefore 1.20 x molecules H 2 : 6.02 x molecules O 2 2. Iron reacts with chlorine gas to form iron(iii) chloride, FeCl 3. 2Fe(s) + 3Cl 2 (g) 2FeCl 3 (s) (a) How many atoms of Fe are needed to react with three molecules of Cl 2? 2 atoms (b) How many formula units of FeCl 3 are formed when 150 atoms of Fe react with sufficient Cl 2? 150 formula units (Ratio of 2:2) (c) How many Cl 2 molecules are needed to react with x atoms of Fe? x atoms Fe x 3 molecules Cl 2 = x molecules Cl 2 2 Atoms Fe ( multiply value given by 3/2 ratio) (d) How many formula units of FeCl 3 are formed when x molecules of Cl 2 react with sufficient Fe? x molecules of Cl 2 x 2 formula units FeCl 3 = x formula units FeCl 3 3 molecules Cl 2 (multiply value given by ratio 2/3) 3. Consider the following reaction. Ca(OH) 2(aq) + 2HCl (aq) CaCl 2 + 2H 2 O (l)

2 (a) How many formula units of calcium chloride, CaCl 2, would be produced by 6.7 x molecules of hydrochloric acid, HCl? 2 molecules HCl : 1 formula unit CaCl 2, (divide by 2), therefore 6.7 x molecules HCl : 3.35 x formula units CaCl 2 = 3.4 x formula units CaCl 2 (b) How many molecules of water would be produced in the reaction in part (a)? 2 molecules HCl : 2 molecules H 2 O, (same ratio), therefore 6.7 x molecules HCl : 6.7 x molecules H 2 O Solutions for Practice Problems p 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. 2Al (s) + 3Br 2(l) 2AlBr 3(s) How many moles of Br 2 are needed to produce 5 mol of AlBr 3, if sufficient Al is present? 5 mol AlBr 3 x 3 mol Br 2 = 7.5 mol Br 2 = 8 mol Br 2 2 mol AlBr 3 5. Hydrogen cyanide gas, HCN(g), is used to prepare clear, hard plastics, such as Plexiglas. Hydrogen cyanide is formed by reacting ammonia, NH 3, with oxygen and methane, CH 4. 2NH 3 (g) + 3O 2 (g) + 2CH 4 (g) 2HCN(g) + 6H 2 O(g) (a) How many moles of O 2 are needed to react with 1.2 mol of NH 3? (b) How many moles of H 2 O can be expected from the reaction of 12.5 mol of CH 4? Assume that sufficient NH 3 and O 2 are present. (a) 1.2 mol of NH 3 x 3 mol O 2 = 1.8 mol O 2 2 mol NH 3 (b) 12.5 mol CH 4 x 6 mol H 2 O = 37.5 mol H 2 O 2 mol CH 4 6. Ethane gas, C 2 H 6, is present in small amounts in natural gas. It undergoes complete combustion to produce carbon dioxide and water. 2C 2 H 6(g) + 7O 2(g) 4CO 2(g) + 6H 2 O (l) (a) How many moles of O 2 are required to react with 13.9 mol of C 2 H 6? 13.9 mol C 2 H 6 x 7 mol O 2 = mol O 2 = 48.6 mol O 2 2 mol C 2 H 6 (b) How many moles of H 2 Owould be produced by 1.40 mol of O 2 and sufficient ethane? 1.40 mol of O 2 x 6 mol H 2 O = 1.20 mol O 2 7 mol O 2

3 7. Magnesium nitride reacts with water to produce magnesium hydroxide and ammonia gas, NH 3 according to the balanced chemical equation Mg 3 N 2 (s) + 6H 2 O(l) 3Mg(OH) 2 (s) + 2NH 3 (g) (a) How many molecules of water are required to react with 2.3 mol Mg 3 N 2? (b) How many formula units of Mg(OH) 2 will be expected in part (a)? (a) 2.3 mol Mg 3 N 2 x 6 mol H 2 O = 13.8 mol H 2 O 1 mol Mg 3 N mol H 2 O x 6.02 x molecules H 2 O = 8.31 x molecules H 2 O 1 mol H 2 O = 8.3 x molecules H 2 O (b) 2.3 mol Mg 3 N 2 x 3 mol Mg(OH) 2 = 6.9 mol Mg(OH) 2 1 mol Mg 3 N mol Mg(OH) 2 x 6.02 x formula units Mg(OH) 2 = 4.2 x fu Mg(OH) 2 1 mol Mg(OH) 2 Solutions for Practice Problems p Ammonium sulfate, (NH 4 ) 2 SO 4, is used as a source of nitrogen in some fertilizers. It reacts with sodium hydroxide to produce sodium sulfate, water, and ammonia. (NH 4 ) 2 SO 4 (s) + 2NaOH (aq) Na 2 SO 4(aq) + 2NH 3(g) + 2H 2 O (l) What mass of sodium hydroxide is required to react completely with 15.4 g of (NH 4 ) 2 SO 4? Solution: m (NH 4 ) 2 SO 4 = 15.4 g m (NaOH) =? Step 1 Convert the given mass (m) of ammonium sulfate, (NH 4 ) 2 SO 4 to the number of moles (n) of ammonium sulfate, using its molar mass (M) g (NH 4 ) 2 SO 4 x 1 mol (NH 4 ) 2 SO 4 [2(14.01) + 8(1.01) (16.00)] g (NH 4 ) 2 SO 4 = 15.4 mol (NH 4 ) 2 SO 4 = mol (NH 4 ) 2 SO Step 2 Determine the mole ratio of NaOH to (NH 4 ) 2 SO 4 from the balanced equation. Using the result in Step 1, solve for the number of moles of NaOH.

4 Mole ratio: mol (NH 4 ) 2 SO 4 x 2 mol NaOH = mol NaOH 1 mol (NH 4 ) 2 SO 4 Step 3 Convert the number of moles of NaOH to mass using its molar mass. ( mol NaOH)( ) g/mol = 9.32 g NaOH 12. 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 2 O 3 (s) + 6HCl (aq) 2FeCl 3(aq) + 3H 2 O (l) What mass of hydrogen chloride is required to react with 1.00 x 10 2 g of rust? Solution: m (Fe 2 O 3 ) = 1.00 x 10 2 g m (HCl) =? (i) 1.00 x 10 2 g Fe 2 O 3 x 1 mol Fe 2 O 3 2(55.85) + 3(16.00)] g Fe 2 O 3 = 1.00 x 10 2 mol = mol Fe 2 O 3 = mol Fe 2 O (ii) mol Fe 2 O 3 x 6 mol HCl = mol HCl 1 mol Fe 2 O 3 (iii) mol HCl x ( ) g HCl = g HCl = 137 g HCl 1 mol HCl 13. Iron reacts slowly with hydrochloric acid to produce iron(iii) chloride and hydrogen gas. Fe (s) + 2HCl (aq) FeCl 2(aq) + H 2(g) What mass of HCl is required to react with 3.56 g of iron? Problem: m (Fe) = 3.56 g m (HCl) =? (i) 3.56 g Fe x 1 mol Fe = mol Fe g Fe (ii) mol Fe x 2 mol HCl = mol HCl 1 mol Fe (iii) mol HCl x g HCl = g HCl = 4.65 g HCl 1 mol HCl

5 Solution for Practice Problem p 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 32.0 g of S 8 reacts with sufficient zinc? Solution: m(s 8 ) = 32.0g m (ZnS) =? (i) 32.0g S 8 x 1 mol = 32.0g mol = mol S 8 8(32.07 g) g (ii) mol S 8 x 8 mol ZnS = mol ZnS 1 mol S 8 (iii) mol ZnS x g g ZnS = ( )(97.46 g ZnS) 1 mol ZnS = g ZnS = 97.2 g ZnS Solutions for Practice Problems p Nitrogen gas is produced in an automobile air bag. It is generated by the decomposition of sodium azide, NaN 3. 2NaN 3(s) 3N 2(g) + 2Na (s) (a) To inflate the air bag on the driver s side of a certain car, 80.0 g of N 2 is required. What mass of NaN 3 is needed to produce 80.0 g of N 2? i) 80.0 g of N 2 x 1 mol N 2 = mol N g N 2 ii) mol N 2 x 2 mol NaN 3 = mol NaN 3 3 mol N 2 iii) mol NaN 3 x g NaN 3 = 124 g NaN 3 1 mol NaN 3 (b) How many atoms of Na are produced when 80.0 g of N 2 are generated in this reaction? i) The number of moles of Na would be equivalent to the number of moles of NaN 3 (ratio of 2:2 in balanced equation). Therefore there would be moles of Na produced. ii) mol Na x 6.02 x atoms Na 1 mol Na = x = 1.14 x atoms Na

6 20. The reaction of iron(iii) oxide with powered aluminum is known as the thermite reaction. 2Al (s) + Fe 2 O 3(s) Al 2 O 3 (s) + 2Fe (s) (a) Calculate the mass of aluminum oxide, Al 2 O 3, that is produced when 1.42 X atoms of Al react with Fe 2 O X atoms Al x 1 mol Al = mol Al 6.02 x atoms mol Al x 1 mol Al 2 O 3 = mol Al 2 O 3 2 mol Al mol Al 2 O 3 x g Al 2 O 3 =120.3 g Al 2 O 3 = 1.20 x 10 2 g Al 2 O 3 1mol Al 2 O 3 (b) How many formula units of Fe 2 O 3 are needed to react with g of Al? g Al x 1 mol Al = mol Al g Al mol Al x 1 mol Fe 2 O 3 = mol Al mol Fe 2 O 3 x 6.02 x formula units Fe 2 O 3 1 mol Fe 2 O 3 = x formula units Fe 2 O 3 = 1.49 x formula units Fe 2 O 3

7 Solutions for Practice Problems p The following balanced chemical equation shows the reaction of aluminum with copper(ii) chloride. If 0.25 g of aluminum reacts with 0.51 g of copper(ii) chloride, determine the limiting reactant. 2Al (s) + 3CuCl 2(aq) 3Cu (s) + 2AlCl 3(aq) Solution: Reactant Al = 0.25 g Reactant CuCl 2 = 0.51 g Convert the given masses into moles. From the balanced equation, identify the coefficient that you will use to divide each quantity of moles, for comparison pursposes only (in order to identify the limiting reactant (LR) (i) 0.25 g Al x 1 mol Al = mol Al g Al mol Al /2 = mol Al 0.51 g CuCl 2 x 1 mol CuCl 2 = 0.51mol CuCl 2 = mol CuCl 2 [ (35.45)] g CuCl The CuCl 2 will run out first so it is the limiting reactant mol CuCl 2 / 3 = mol CuCl Hydrogen fluoride, HF, is a highly toxic gas. It is produced by the double displacement reaction of calcium fluoride, CaF 2, with concentrated sulfuric acid, H 2 SO 4. CaF 2(s) + H 2 SO 4(l) 2HF (g) + CaSO 4(s) Determine the limiting reactant when 10.0 g of CaF 2 reacts with 15.5 g of H 2 SO 4. Solution: Reactant CaF 2 = 10.0 g Reactant H 2 SO 4 = 15.5 g (i) 10.0 g CaF 2 x 1 mol CaF 2 = 10.0 mol CaF 2 = mol CaF 2 [ (19.00)] g CaF g H 2 SO 4 x 1 mol H 2 SO 4 = 15.5 mol H 2 SO 4 [2(1.01) (16.00)] g H 2 SO = mol H 2 SO 4 As the ratio of each reactant is 1:1, we can determine that CaF 2 is the limiting reactant (will run out first).

8 25. Acrylic, a common synthetic fibre, is formed from acrylonitrile C 3 H 3 N. Acrylonitrile can be prepared by the reaction of propylene, C 3 H 6, with nitric oxide, NO. 4C 3 H 6 (g) + 6NO (g) 4C 3 H 3 N (g) + 6H 2 O (g) + N 2(g) What is the limiting reactant when 126 g of C 3 H 6 reacts with 175 g of NO? Solution: Reactant C 3 H 6 = 126 g Reactant NO = 175 g (i) 126 g C 3 H 6 x 1 mol C 3 H 6 = 126 mol C 3 H 6 = mol C 3 H 6 [3( (1.01)] g C 3 H g NO x 1 mol NO = 175 g = mol NO ( ) g NO g/mol (ii) Comparison of amount of moles (using mole ratio): mol C 3 H 6 /4 = mol C 3 H 6 (less so LR) mol NO /6 = mol NO C 3 H 6 is used up first, it is therefore the limiting reactant g of zinc reacts with 8.93 x hydrogen ions as shown in the equation below. Which reactant is present in excess? Zn (s) + 2H + (aq) Zn 2+ (aq) + H 2(g) Solution: Reactant Zn = 3.76 g Reactant H + = 8.93 x ions. N A = 6.02 x ions/mol (i) 3.76 g Zn x 1 mol Zn = mol Zn g Zn 8.93 x ions H + x 1 mol H + = mol H x ions H + (ii) mol H + / 2 = mol H + (which is more than other reactant) H + is present in excess.

9 Solutions for Practice Problems p Chlorine dioxide, ClO 2, is a reactive oxidizing agent. It is used to purify water. 6ClO 2(g) + 3H 2 O (l) 5HClO 3(aq) + HCl (aq) (a) If g of ClO 2 is mixed with g of water, what is the limiting reactant? (b) What mass of HClO 3 expected in part (a)? (c) How many molecules of HCl are expected in part (a)? Given: Mass of reactant ClO 2 = g Mass of reactant H 2 O = g Solution: (a) Determine the # moles of each reactant available, then divide that amount by the coefficient in the balanced equation for comparative purposes in order to identify the limiting reactant (LR) g ClO 2 x 1 mol ClO 2 = mol ClO 2 = mol ClO 2 ( ) g ClO g mol ClO 2 / 6 (coefficient) = mol ClO g H 2 O x 1 mol H 2 O = mol H 2 O = mol H 2 O ( ) g H 2 O mol H 2 O / 3 (coefficient) = mol H 2 O The limiting reactant is ClO 2. (b) Use the initial molar amount of the limiting reactant for future calculations mol ClO 2 x 5 mol HClO 3 = mol HClO 3 6 mol ClO 2 ( mol HClO 3 )( x16.00) g = ( )(84.46 g) = g HClO 3 1 mol (c) Using the number of moles of ClO 2 (the L.R.): mol ClO 2 x 1 mol HCl = mol HCl 6 mol ClO mol HCl x 6.02 x molecules HCl = x molecules HCl mol HCl

10 28. Hydrazine, N 2 H 4, reacts exothermically with hydrogen peroxide, H 2 O 2. N 2 H 4(l) + 7H 2 O 2(aq) 2HNO 3(g) + 8H 2 O (g) (a) 120 g of N 2 H 4 reacts with an equal mass of H 2 O 2. Which is the limiting reactant? (b) What mass of HNO 3 is expected? (c) What mass, in grams, of the excess reactant remains at the end of the reaction? Solution: N 2 H 4(l) + 7H 2 O 2(aq) 2HNO 3(g) + 8H 2 O (g) 120 g 120 g (a) 120 g N 2 H 4 x 1 mol N 2 H 4 = 120 mol N 2 H 4 = mol N 2 H 4 ( ) g N 2 H g H 2 O 2 x 1 mol H 2 O 2 = 120 mol H 2 O 2 = mol H 2 O 2 ( ) g H 2 O g H 2 O 2 The limiting reactant is H 2 O 2. This answer may be supported in one of the following ways: i) The number of moles of H 2 O 2 is less than 7 times the amount it should be as shown in the above balanced chemical equation. This means that H 2 O 2 is used up first. ii) mol H 2 O 2 x 1 mol N 2 H 4 = mol N 2 H 4 7 mol H 2 O 2 A quantity which is smaller than the calculated availability of N 2 H 4. Therefore H 2 O 2 is used up first (b) Use LR (limiting reactant) for future calculations mol H 2 O 2 x 2 mol HNO 3 = mol HNO 3 7 mol H 2 O mol HNO 3 x ( x16.00) g = (1.008)(63.02 g HNO 3 ) = g HNO 3 1 mol HNO 3 = 64 g HNO 3 (c) Using the mole ratio of the limiting reactant, one can determine the amount of N 2 H 4 that was used up. Subtract the amount used from the mole amount calculated in (a) for the excess reactant. Multiply the mole difference by the molar mass of the excess reactant to obtain the mass left over. (i) mol H 2 O 2 x 1 mol N 2 H 4 = mol N 2 H 4 used up 7 mol H 2 O 2 (ii)3.743 mol N 2 H 4 available mol N 2 H 4 used = mol N 2 H 4 in excess (iii) (3.239 mol N 2 H 4 ) (2 x x 1.01) g N 2 H 4 = (3.239)(32.06 g N 2 H 4 ) 1 mol N 2 H 4 = g N 2 H 4 = 1.0 x 10 2 g N 2 H 4 remains at the end of the reaction

11 29. 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 2 S 2 O 3(aq). Na 2 S 2 O 3(aq) + 4Cl 2(g) + 5H 2 O (l) 2NaHSO 4(aq) + 8HCl (aq) 135 kg of Na 2 S 2 O 3 reacts with 50.0 kg of Cl 2 and 238 kg of water. How many grams of NaHSO 4 are expected to be produced? What Is Given? Remember to convert the given kg quantities to grams before proceeding with the calculations. Na 2 S 2 O 3(aq) + 4Cl 2(g) + 5H 2 O (l) 2NaHSO 4(aq) + 8HCl (aq) Reactant Na 2 S 2 O 3 = 135 kg = g Reactant Cl 2 = 50.0 kg = 5.00 x 10 4 g Reactant H 2 O = 238 kg = g Step 1: Determine the limiting reactant: (i) g Na 2 S 2 O 3 x 1 mol Na 2 S 2 O 3 = mol Na 2 S 2 O 3 [2(22.99) + 2(32.07) + 3(16.00)] g = mol Na 2 S 2 O 3 (For comparison purposes only so as to identify L.R., divide # mol by coefficient in balanced equation above) mol Na 2 S 2 O 3 / 1 = mol Na 2 S 2 O 3 (ii) 5.00 x 10 4 g Cl 2 x 1 mol Cl 2 = 5.00 x 10 4 mol Cl 2 = mol Cl 2 2 (35.45) g Cl (For comparison purposes only so as to identify L.R., divide # mol by coefficient in balanced equation above) mol Cl 2 / 4 = mol Cl 2 (iii) g H 2 O x 1 mol H 2 O = mol H 2 O g H 2 O (For comparison purposes only so as to identify L.R., divide # mol by coefficient in balanced equation above) mol H 2 O / 5 = mol H 2 O The limiting reactant is chlorine (Cl 2 ) as it is the reactant that will run out first. [ mol Na 2 S 2 O 3 : mol Cl 2 : mol H 2 O] Use the number of moles *available* of the LR for any remaining calculations. Step 2: Cacu1ate the number of moles of NaHSO 4 produced by the limiting reactant mol Cl 2 x 2 mol NaHSO 4 = mol NaHSO 4 4 mol Cl 2 Step 3: Calculate the mass of NaHSO 4 ( mol NaHSO 4 )( ) g = (352.61)( g NaHSO 4 ) 1 mol NaHSO 4 = g NaHSO 4 = 4.23 x 10 4 g NaHSO 4

12 30. Manganese(III) fluoride can be formed by the reaction of manganese(ii) iodide with fluorine. 2MnI 2(s) + 13F 2(g) 2MnF 3(s) + 4IF 5(l) (a) 1.23 g of MnI 2 reacts with 25.0 g of F 2. What mass of MnF 3 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? Solution: Reactant MnI 2 = 1.23 g Reactant F 2 = 25.0 g (a) Determine the identity of the limiting reactant for the given amounts of reactants. (i) 1.23 g MnI 2 x 1 mol MnI 2 = 1.23 mol MnI 2 = mol MnI 2 ( x ) g mol MnI 2 /2 (coefficient) = mol MnI 2 (ii) 25.0 g F 2 x 1 mol F 2 = mol F mol F 2 / 13 (coefficient) = mol F 2 MnI 2 is the limiting reactant as it will run out before F 2 Use the original amount of moles of limiting reactant as starting point for all future calculations. (iii) mol MnI 2 x 2 mol MnF 3 = mol MnF 3 2 mol MnI 2 ( mol MnF 3 )( x 19.00) g 1 mol MnF 3 = ( mol MnF 3 )( g/mol MnF 3 ) = g = g MnF 3 (b) (i) mol MnI 2 x 4 mol IF 5 = mol IF 5 2 mol MnI 2 (ii) ( mol IF 5 )(6.02 x molecules IF 5 ) = 4.80 x molecules IF 5 mol IF 5 (c) Use the mole ratio of the limiting reactant to determine the number of moles of the excess reactant, F 2. Subtract this value from the initial calculated molar amount of F 2. (i) Calculate moles of excess reactant used up: mol MnI 2 x 13 mol F 2 = mol F 2 used 2 mol MnI 2 (ii) mol F 2 available mol F 2 used = mol F 2 in excess (iii) ( mol)(38.00 g/mol) = 24.0 g excess F 2

13 Solutions for Practice Problems p g of bromic acid, HBrO 3, are reacted with excess HBr. HBrO 3(aq) + 5HBr (aq) 3H 2 O (l) + 3Br 2 (aq) (a) What is the theoretical yield of Br 2 for this reaction? 20.0 g HBrO 3 x 1 mol HBrO 3 = 20.0 mol HBrO 3 [ (16.00)] g HBrO = mol HBrO mol HBrO 3 x 3 mol Br 2 = mol Br 2 1 mol HBrO mol Br 2 x 2(79.90g Br 2 ) = x 159.8g = g Br 2 = 74.4 g Br 2 1 mol Br 2 (b) If 47.3 g of Br 2 are produced, what is the percentage yield of Br 2? Known info: Actual yield = 47.3 g Br 2 Theoretical yield calculated in (a) = 74.4 g Br 2 % yield = Actual yield x 100% = 47.3 g Br 2 x 100% = x 100% = 63.6 % Theoretical yield 74.4 g Br Barium sulfate forms as a precipitate in the following reaction: Ba(NO 3 ) 2(aq) + NaSO 4(aq) BaSO 4(s) + 2NaNO 3(aq) When 35.0 g of Ba(NO 3 ) 2 are reacted with excess NaSO 4, 29.8 g of BaSO 4 is recovered by the chemist. (a) Calculate the theoretical yield of BaSO 4. (i) 35.0 g Ba(NO 3 ) 2 x 1 mol Ba(NO 3 ) 2 [ (14.01) + 6(16.00)] g 35.0 g Ba(NO 3 ) 2 x 1 mol Ba(NO 3 ) 2 = mol Ba(NO 3 ) 2 [261.35] g Ba(NO 3 ) 2 (ii) mol Ba(NO 3 ) 2 x 1 mol BaSO 4 = mol BaSO 4 1 mol Ba(NO 3 ) 2 (iii) mol BaSO 4 x (16.00) 1 mol BaSO 4 = mol BaSO 4 x g BaSO 4 = g BaSO 4 = 31.2 g BaSO 4 1 mol BaSO 4

14 (b) Calculate the percentage yield of BaSO 4. Known info: Actual yield = 29.8 g BaSO 4 recovered Theoretical yield calculated in (a) = 31.3 g BaSO 4 % yield = Actual yield x 100% = 29.8 g Br 2 x 100% = x 100% = 95.5 % Theoretical yield 31.2 g Br Yeasts can act on a sugar, such as glucose, C 6 H 12 O 6, to produce ethyl alcohol, C 2 H 5 OH, and carbon dioxide: C 6 H 12 O 6 2 C 2 H 5 OH + 2CO 2 If 223 g of ethyl alcohol are recovered after 1.63 kg of glucose react, what is the percentage yield of the reaction? Known info: Reactant glucose = 1.63 kg = 1630 g Actual yield of ethyl alcohol = 223 g. The balanced equation is given. i) 1630 g C 6 H 12 O 6 x 1 mol C 6 H 12 O 6 [6(12.01) + 12(1.01) + 6(16.00)] g 1630 g C 6 H 12 O 6 x 1 mol C 6 H 12 O 6 = mol C 6 H 12 O g C 6 H 12 O mol C 6 H 12 O 6 x 2 mol C 2 H 5 OH = mol C 2 H 5 OH 1 mol C 6 H 12 O 6 ii) mol C 2 H 5 OH x g C 2 H 5 OH = g C 2 H 5 OH = 834 g 1 mol C 2 H 5 OH iii) Actual yield of ethyl alcohol = 223 g. Theoretical yield calculated above is 834 g. % yield = Actual yield x 100% = 223 g Br 2 x 100% = x 100% = 26.7 % Theoretical yield 834 g Br 2 Solutions for Practice Problems p The following reaction proceeds with a 70% yield. C 6 H 6 (l) HNO 3(aq) C 6 H 5 NO 2(l) H 2 O (l)

15 Calculate the mass of C 6 H 5 NO 2 expected if 12.8 g of C 6 H 6 expected if 12.8 g of C 6 H 6 reacts with excess HNO 3. Snapshot: 12.8 g C 6 H 6 Percentage yield of C 6 H 5 NO 2 is 70%. Actual yield (g) =? Solution: i) First find the theoretical yield (g) 12.8 g of C 6 H 6 x 1 mol = 12.8 mol = mol C 6 H 6 [6(12.01) + 6(1.01)]g mol C 6 H 6 x 1 mol C 6 H 5 NO 2 = mol C 6 H 5 NO 2 1 mol C 6 H mol C 6 H 5 NO 2 x [6(12.01) + 5(1.01) (16.00)]g 1 mol = x g = g = 20.2 g C 6 H 5 NO 2 ii) Convert the percentage yield to a decimal. Then multiply it by the theoretical yield x 20.2 g = 14.1 g Actual Yield

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