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Chapter 9 Chemical Quantities 1. Although we define mass as the “amount of matter in a substance,” the units in which we measure mass are a human invention. Atoms and molecules react on an individual particleby-particle basis, and we have to count individual particles when doing chemical calculations. 2. Balanced chemical equations tell us in what proportions on a mole basis substances combine; since the molar masses of C(s) and O2(g) are different, 1 g of O2 could not represent the same number of moles as 1 g of C. 3. a. 2NO(g) + O2(g) 2NO2(g) Two molecules of nitrogen monoxide combine with one molecule of oxygen gas, producing two molecules of nitrogen dioxide. Two moles of gaseous nitrogen monoxide combine with one mole of gaseous oxygen, producing two moles of gaseous nitrogen dioxide. b. 2AgC2H3O2(aq) + CuSO4(aq) Ag2SO4(s) + Cu(C2H3O2)2(aq) Note: The term “formula unit” is used in the following statement because the substances involved in the above reaction are ionic, and do not contain true molecules. Two formula units of silver acetate will react with one formula unit of copper(II) sulfate, precipitating one formula unit of silver sulfate and leaving one formula unit of copper(II) acetate in solution. Two moles of aqueous silver acetate react with one mole of aqueous copper(II) sulfate, to produce one mole of solid silver sulfate as a precipitate, and leaving one mole of copper(II) acetate in solution. c. PCl3(l) + 3H2O(l) H3PO3(l) + 3HCl(g) One molecule of phosphorus trichloride reacts with three molecules of water, producing one molecule of phosphorous acid and three molecules of gaseous hydrogen chloride. One mole of liquid phosphorus trichloride reacts with three moles of liquid water, producing one mole of liquid phosphorous acid and three moles of gaseous hydrogen chloride. d. C2H6(g) + Cl2(g) C2H5Cl(g) + HCl(g) One molecule of ethane (C2H6) reacts with one molecule of chlorine, producing one molecule of chloroethane (C2H5Cl) and one molecule of hydrogen chloride. One mole of gaseous ethane combines with one mole of chlorine gas, giving one mole of gaseous chloroethane and one mole of gaseous hydrogen chloride. 4. a. 3MnO2(s) + 4Al(s) 3Mn(s) + 2Al2O3(s) Three formula units of manganese(IV) oxide react with four aluminum atoms, producing three manganese atoms and two formula units of aluminum oxide. Three moles of solid manganese(IV) oxide react with four moles of solid aluminum, to produce three moles of solid manganese and two moles of solid aluminum oxide. World of Chemistry 74 Copyright Houghton Mifflin Company. All rights reserved. Chemical Quantities b. 75 B2O3(s) + 3CaF2(s) 2BF3(g) + 3CaO(s) One molecule of diboron trioxide reacts with three formula units of calcium fluoride, producing two molecules of boron trifluoride and three formula units of calcium oxide. One mole of solid diboron trioxide reacts with three moles of solid calcium fluoride, to give two moles of gaseous boron trifluoride and three moles of solid calcium oxide. c. 3NO2(g) + H2O(l) 2HNO3(aq) + NO(g) Three molecules of nitrogen dioxide [nitrogen(IV) oxide] react with one molecule of water, to produce two molecules of nitric acid and one molecule of nitrogen monoxide [nitrogen(II) oxide]. Three moles of gaseous nitrogen dioxide react with one mole of liquid water, to produce two moles of aqueous nitric acid and one mole of nitrogen monoxide gas. d. C6H6(g) + 3H2(g) C6H12(g) One molecule of C6H6 (which is named benzene) reacts with three molecules of hydrogen, producing just one molecule of C6H12 (which is named cyclohexane). One mole of gaseous benzene reacts with three moles of hydrogen gas, giving one mole of gaseous cyclohexane. 5. False. The coefficients of the balanced chemical equation represent the ratios on a mole basis by which hydrogen peroxide decomposes. 6. For converting from a given number of moles of CH4 to the number of moles of oxygen needed for reaction, the correct mole ratio is 2 mol O 2 1 mol CH 4 For converting from a given number of moles of CH4 to the number of moles of product produced, the ratios are 1 mol CO 2 1 mol CH 4 7. 2 mol H 2 O 1 mol CH 4 2Ag(s) + H2S(g) Ag2S(s) + H2(g) 1 mol Ag 2S 2 mol Ag 8. and and 1 mol H 2 2 mol Ag a. 2FeO(s) + C(s) 2Fe(l) + CO2(g) 0.125 mol FeO 0.125 mol FeO World of Chemistry 2 mol Fe 2 mol FeO 1 mol CO 2 2 mol FeO = 0.125 mol Fe = 0.0625 mol CO2 Copyright Houghton Mifflin Company. All rights reserved. 76 Chapter 9 b. Cl2(g) + 2KI(aq) 2KCl(aq) + I2(s) 0.125 mol KI 0.125 mol KI c. 2 mol KCl 1 mol I 2 2 mol KI 0.125 mol Na2B4O7 0.125 mol CaC2 a. 4 mol H 3 BO3 1 mol Na 2 B4 O7 1 mol Na 2SO 4 1 mol Na 2 B4 O7 = 0.500 mol H3BO3 = 0.125 mol Na2SO4 CaC2(s) + 2H2O(l) Ca(OH)2(s) + C2H2(g) 0.125 mol CaC2 9. = 0.0625 mol I2 Na2B4O7(s) + H2SO4(aq) + 5H2O(l) 4H3BO3(s) + Na2SO4(aq) 0.125 mol Na2B4O7 d. = 0.125 mol KCl 2 mol KI 1 mol Ca(OH) 2 1 mol CaC 2 1 mol C 2 H 2 1 mol CaC 2 = 0.125 mol Ca(OH)2 = 0.125 mol C2H2 NH3(g) + HCl(g) NH4Cl(s) molar mass of NH4Cl, 53.49 g 0.50 mol NH3 1 mol NH 4 Cl 1 mol NH 3 0.50 mol NH4Cl b. = 0.50 mol NH4Cl 53.49 g NH 4 Cl 1 mol NH 4 Cl = 27 g NH4Cl CH4(g) + 4S(g) CS2(l) + 2H2S(g) molar masses: CS2, 76.15 g; H2S, 34.09 g 0.50 mol S 1 mol CS2 4 mol S 0.125 mol CS2 0.50 mol S c. 76.15 g CS2 1 mol CS2 2 mol H 2S 4 mol S 0.25 mol H2S = 0.125 mol CS2 (= 0.13 mol CS2) = 9.5 g CS2 = 0.25 mol H2S 34.09 g H 2S 1 mol H 2S = 8.5 g H2S PCl3(l) + 3H2O(l) H3PO3(aq) + 3HCl(aq) World of Chemistry Copyright Houghton Mifflin Company. All rights reserved. Chemical Quantities 77 molar masses: H3PO3, 81.99 g; HCl, 36.46 g 0.50 mol PCl3 1 mol H 3 PO3 81.99 g H 3 PO3 0.50 mol H3PO3 0.50 mol PCl3 1.50 mol HCl d. = 0.50 mol H3PO3 1 mol PCl3 = 41 g H3PO3 1 mol H 3 PO3 3 mol HCl 1 mol PCl3 = 1.5 mol HCl 36.46 g HCl 1 mol HCl = 54.7 = 55 g HCl NaOH(s) + CO2(g) NaHCO3(s) molar mass of NaHCO3, 84.01 g 1 mol NaHCO3 0.50 mol NaOH 1 mol NaOH 0.50 mol NaHCO3 10. = 0.50 mol NaHCO3 84.01 g NaHCO3 1 mol NaHCO3 = 42 g NaHCO3 Before doing the calculations, the equations must be balanced. a. 4KO2(s) + 2H2O(l) 3O2(g) + 4KOH(s) 3 mol O 2 0.625 mol KOH b. 3 mol Se 2 mol H 2 O = 0.938 mol Se 2CH3CH2OH(l) + O2(g) 2CH3CHO(aq) + 2H2O(l) 0.625 mol H2O d. = 0.469 mol O2 SeO2(g) + 2H2Se(g) 3Se(s) + 2H2O(g) 0.625 mol H2O c. 4 mol KOH 2 mol CH 3CHO 2 mol H 2 O = 0.625 mol CH3CHO Fe2O3(s) + 2Al(s) 2Fe(l) + Al2O3(s) 0.625 mol Al2O3 2 mol Fe 1 mol Al2 O3 = 1.25 mol Fe 11. the molar mass of the substance 12. Stoichiometry is the process of using a chemical equation to calculate the relative masses of reactants and products involved in a reaction. World of Chemistry Copyright Houghton Mifflin Company. All rights reserved. 78 13. Chapter 9 a. molar mass of Ag = 107.9 g 1 mol 2.01 10–2 g Ag b. molar mass of (NH4)2S = 68.15 g 45.2 mg (NH4)2S c. 10 g a. 1 kg = 81.6 mol SO2 1 mol 241.9 g = 1.12 mol Fe(NO3)3 1g 1000 mg 1 mol 151.9 g = 8.36 10–5 mol FeSO4 molar mass of LiOH = 23.95 g 1 mol 23.95 g = 288.5 = 289 mol LiOH molar mass of CaCO3 = 100.1 g 100.1 g 1 mol = 0.0221 g CaCO3 molar mass of He = 4.003 g 4.003 g 1 mol = 11.0 g He molar mass of O2 = 32.00 g 0.00975 mol O2 d. 64.07 g molar mass of FeSO4 = 151.9 g 2.75 mol He c. 1 mol molar mass of Fe(NO3)3 = 241.9 g 2.21 10–4 mol CaCO3 b. = 6.63 10–4 mol (NH4)2S = 2.59 10–7 mol U 238.0 g 1000 g 6.91 103 g LiOH 14. 68.15 g molar mass of SO2 = 64.07 g 12.7 mg FeSO4 g. 1 mol 1 mol 6 272 g Fe(NO3)3 f. 1000 mg 1g 5.23 kg SO2 e. 1g molar mass of uranium = 238.0 g 61.7 g U d. = 1.86 10–4 mol Ag 107.9 g 32.00 g 1 mol = 0.312 g O2 molar mass of CO2 = 44.01 g 7.21 millimol = 0.00721 mol 0.00721 mol CO2 World of Chemistry 44.01 g 1 mol = 0.317 g CO2 Copyright Houghton Mifflin Company. All rights reserved. Chemical Quantities e. 79 molar mass of FeS = 87.92 g 87.92 g 0.835 mol FeS f. molar mass of KOH = 56.11 g 56.11 g 4.01 mol KOH g. = 225 g KOH 1 mol molar mass of H2 = 2.016 g 2.016 g 0.0219 mol H2 15. = 73.4 g FeS 1 mol = 0.0442 g H2 1 mol Before any calculations are done, the equations must be balanced. a. Mg(s) + CuCl2(aq) MgCl2(aq) + Cu(s) molar mass of Mg = 24.31 g 25.0 g Mg 1 mol 1.03 mol Mg b. = 1.03 mol Mg 24.31 g 1 mol CuCl2 1 mol Mg = 1.03 mol CuCl2 2AgNO3(aq) + NiCl2(aq) 2AgCl(s) + Ni(NO3)2(aq) molar mass of AgNO3 = 169.9 g 25.0 g AgNO3 1 mol 169.9 g 0.147 mol AgNO3 c. = 0.147 mol AgNO3 1 mol NiCl2 2 mol AgNO3 = 0.0735 mol NiCl2 NaHSO3(aq) + NaOH(aq) Na2SO3(aq) + H2O(l) molar mass of NaHSO3 = 104.1 g 25.0 g NaHSO3 1 mol 104.1 g 0.240 mol NaHSO3 d. = 0.240 mol NaHSO3 1 mol NaOH 1 mol NaHSO3 = 0.240 mol NaOH KHCO3(aq) + HCl(aq) KCl(aq) + H2O(l) + CO2(g) molar mass of KHCO3 = 100.1 g 25.0 g KHCO3 1 mol 100.1 g 0.250 mol KHCO3 World of Chemistry = 0.250 mol KHCO3 1 mol HCl 1 mol KHCO3 = 0.250 mol HCl Copyright Houghton Mifflin Company. All rights reserved. 80 16. Chapter 9 Before any calculations are done, the equations must be balanced. Since the given and required quantities in this question are shown in milligrams, it is most convenient to perform the calculations in terms of millimoles of the substances involved. One millimole of a substance represents the molar mass of the substance expressed in milligrams. a. FeSO4(aq) + K2CO3(aq) FeCO3(s) + K2SO4(aq) millimolar masses: FeSO4, 151.9 mg; FeCO3, 115.9 mg; K2SO4, 174.3 mg 1 mmol FeSO 4 10.0 mg FeSO4 0.0658 mmol FeSO4 0.0658 mmol FeSO4 b. = 0.0658 mmol FeSO4 151.9 mg FeSO 4 1 mmol FeCO3 1 mmol FeSO 4 1 mmol K 2SO 4 1 mmol FeSO 4 115.9 mg FeCO3 1 mmol FeCO3 174.3 mg K 2SO 4 1 mmol K 2SO 4 = 7.63 mg FeCO3 = 11.5 mg K2SO4 4Cr(s) + 3SnCl4(l) 4CrCl3(s) + 3Sn(s) millimolar masses: Cr, 52.00 mg; CrCl3, 158.4 mg; Sn, 118.7 mg 10.0 mg Cr 1 mmol Cr 52.00 mg Cr 4 mmol CrCl3 0.192 mmol Cr 4 mmol Cr 3 mmol Sn 0.192 mmol Cr c. = 0.192 mmol Cr 4 mmol Cr 158.4 mg CrCl3 1 mmol CrCl3 118.7 mg Sn 1 mmol Sn = 30.4 mg CrCl3 = 17.1 mg Sn 16Fe(s) + 3S8(s) 8Fe2S3(s) millimolar masses: S8, 256.6 mg; Fe2S3, 207.9 mg 10.0 mg S8 1 mmol S8 256.6 mg S8 8 mmol Fe 2S3 0.0390 mmol S8 d. = 0.0390 mmol S8 3 mmol S8 207.9 mg Fe 2S3 1 mmol Fe 2S3 = 21.6 mg Fe2S3 3Ag(s) + 4HNO3(aq) 3AgNO3(aq) + 2H2O(l) + NO(g) millimolar masses: HNO3, 63.0 mg; AgNO3, 169.9 mg H2O, 18.0 mg; NO, 30.0 mg 10.0 mg HNO3 1 mmol HNO3 63.0 mg HNO3 0.159 mmol HNO3 World of Chemistry = 0.159 mmol HNO3 3 mmol AgNO 3 4 mmol HNO 3 169.9 mg AgNO 3 1 mmol AgNO 3 = 20.3 mg AgNO3 Copyright Houghton Mifflin Company. All rights reserved. Chemical Quantities 81 0.159 mmol HNO3 0.159 mmol HNO3 17. 2 mmol H 2 O 4 mmol HNO 3 1 mmol NO 4 mmol HNO 3 18.0 mg H 2 O 1 mmol H 2 O 30.0 mg NO 1 mmol NO = 1.43 mg H2O = 1.19 mg NO 2H2(g) + O2(g) 2H2O(g) molar masses: H2, 2.016 g; H2O, 18.02 g 56.0 g H2 1 mol H 2 2.016 g H 2 27.77 mol H2 2 mol H 2 O = 27.77 mol H2O 2 mol H 2 18.02 g H 2 O 27.77 mol H2O 18. = 27.77 mol H2 1 mol H 2 O = 500. g H2O 2H2(g) + O2(g) 2H2O(g) molar masses of O2 = 32.00 g 1 mol O 2 0.0275 mol H2 2 mol H 2 32.00 g O 2 0.01375 mol O2 19. = 0.01375 = 0.0138 mol O2 1 mol O 2 = 0.440 g O2 molar masses: C, 12.01 g; CO, 28.01 g; CO2, 44.01 g 5.00 g C 1 mol C 12.01 g C = 0.4163 mol C carbon dioxide: C(s) + O2(g) CO2(g) 0.4163 mol C 1 mol CO 2 = 0.4163 mol CO2 1 mol C 0.4163 mol CO2 44.01 g CO 2 1 mol CO 2 = 18.3 g CO2 carbon monoxide: 2C(s) + O2(g) 2CO(g) 0.4163 mol C 2 mol CO 0.4163 mol CO World of Chemistry 2 mol C = 0.4163 mol CO 28.01 g CO 1 mol CO = 11.7 g CO Copyright Houghton Mifflin Company. All rights reserved. 82 20. Chapter 9 2Fe(s) + 3Cl2(g) 2FeCl3(s) millimolar masses: Fe, 55.85 mg; FeCl3, 162.2 mg 15.5 mg Fe 1 mmol Fe 55.85 mg Fe 2 mmol FeCl3 0.2775 mmol Fe = 0.2775 mmol FeCl3 2 mmol Fe 0.2775 mmol FeCl3 21. = 0.2775 mmol Fe 162.2 mg FeCl3 1 mmol FeCl3 = 45.0 mg FeCl3 2H2O2(aq) 2H2O(l) + O2(g) molar masses: H2O2, 34.02 g; O2, 32.00 g 10.00 g H2O2 1 mol H 2 O 2 0.2939 mol H2O2 0.1470 mol O2 22. = 0.2939 mol H2O2 34.02 g H 2 O 2 1 mol O 2 2 mol H 2 O 2 32.00 g O 2 = 0.1470 mol O2 = 4.704 g O2 1 mol O 2 Cu(s) + S(s) CuS(s) molar masses: Cu, 63.55 g; S, 32.07 g 1.25 g Cu 1 mol = 1.97 10–2 mol Cu 63.55 g 1.97 10–2 mol Cu 1.97 10–2 mol S 23. 1 mol S 1 mol Cu 32.07 g 1 mol = 1.97 10–2 mol S = 0.631 g S 2NH4NO3(s) 2N2(g) + O2(g) + 4H2O(g) molar masses: NH4NO3, 80.05 g; N2, 28.02 g; O2, 32.00 g; H2O, 18.02 g 1.25 g NH4NO3 1 mol NH 4 NO3 80.05 g NH 4 NO3 0.0156 mol NH4NO3 0.0156 mol N2 World of Chemistry = 0.0156 mol NH4NO3 2 mol N 2 2 mol NH 4 NO3 28.02 g N 2 1 mol N 2 = 0.0156 mol N2 = 0.437 g N2 Copyright Houghton Mifflin Company. All rights reserved. Chemical Quantities 83 1 mol O 2 0.0156 mol NH4NO3 0.00780 mol O2 2 mol NH 4 NO3 32.00 g O 2 = 0.250 g O2 1 mol O 2 4 mol H 2 O 0.0156 mol NH4NO3 2 mol NH 4 NO3 18.02 g H 2 O 0.0312 mol H2O = 0.00780 mol O2 = 0.0312 mol H2O = 0.562 g H2O 1 mol H 2 O As a check, note that 0.437 g + 0.250 g + 0.562 g = 1.249 g = 1.25 g. 24. 2Mg(s) + O2(g) 2MgO(s) molar masses: Mg, 24.31 g; MgO, 40.31 g 1.25 g Mg 1 mol 24.31 g = 5.14 10–2 mol Mg 5.14 10–2 mol Mg 2 mol MgO 5.14 10–2 mol MgO 25. = 5.14 10–2 mol MgO 2 mol Mg 40.31 g = 2.07 g MgO 1 mol From the balanced equation: Cl2 + 2KI I2 + 2KCl We can calculate the following: 4.50 103 g Cl2 26. 1 mol Cl2 70.9 g Cl2 1 mol I 2 1 mol Cl2 253.8 g I 2 1 mol I 2 = 1.61 104 g I2 From the balanced equation: Cl2 + F2 2ClF We can calculate the following: 5.00 10–3 g ClF 27. 1 mol ClF 54.45 g ClF 1 mol Cl2 2 mol ClF 70.9 g Cl2 1 mol Cl2 = 3.26 10–3 g Cl2 Start by balancing the equations: C3H8 + 5O2 3CO2 + 4H2O 2C4H10 + 13O2 8CO2 + 10H2O If we had, for example, 10.00 g of C3H8 and 10.00 g of C4H10, we can calculate the mass of oxygen required for each reaction and compare the results. World of Chemistry Copyright Houghton Mifflin Company. All rights reserved. 84 Chapter 9 10.00 g C3H8 10.00 g C4H10 1 mol C3 H 8 44.09 g C3 H 8 1 mol C 4 H 10 5 mol O 2 1 mol C3 H 8 13 mol O 2 58.12 g C 4 H 10 2 mol C 4 H 10 32.00 g O 2 1 mol O 2 = 36.29 g O2 32.00 g O 2 1 mol O 2 = 35.79 g O2 Therefore, more O2 would be required for the combustion of C3H8. The same result is obtained with masses other than 10.00 g. 28. Start with the balanced equations: CH4 + 2O2 CO2 + 2H2O 4NH3 + 5O2 4NO + 6H2O The amount of water produced from 1.00 g CH4 is: 1.00 g CH4 1 mol CH 4 16.042 g CH 4 2 mol H 2 O 1 mol CH 4 = 0.125 mol H2O The mass of NH3 needed to produce 0.125 mol H2O is: 0.125 mol H2O 4 mol NH 3 6 mol H 2 O 17.034 g NH 3 1 mol NH 3 = 1.42 g NH3 29. The limiting reactant is the reactant that limits the amounts of products that can form in a chemical reaction. All given reactants are necessary for the production of products: if the limiting reactant has been consumed, then there is none of this reactant present for reaction. 30. We start with 6 molecules N2 and 6 molecules H2, and these react according to the balanced equation: N2 + 3H2 2NH3 We can determine amounts of product and leftover reactant from this information given. N2 + 3H2 2NH3 start react end 6 6 –2 –6 4 0 0 +4 4 Note that these react in the same ratio as given in the balanced equation (that is, 2:6:4 = 1:3:2). The pictures should show 4 molecules N2 and 4 molecules NH3. 31. To determine the limiting reactant, first calculate the number of moles of each reactant present. Then determine how these numbers of moles correspond to the stoichiometric ratio indicated by the balanced chemical equation for the reaction. 32. The theoretical yield of a reaction represents the stoichiometric amount of product that should form if the limiting reactant for the process is completely consumed. World of Chemistry Copyright Houghton Mifflin Company. All rights reserved. Chemical Quantities 85 33. A reactant is present in excess if there is more of that reactant present than is needed to combine with the limiting reactant for the process. By definition, the limiting reactant cannot be present in excess. An excess of any reactant does not affect the theoretical yield for a process: the theoretical yield is determined by the limiting reactant. 34. a. 2Al(s) + 6HCl(aq) 2AlCl3(aq) + 3H2(g) Molar masses: Al, 26.98 g; HCl, 36.46 g; AlCl3, 133.3 g; H2, 2.016 g 15.0 g Al 1 mol 26.98 g 15.0 g HCl = 0.556 mol Al 1 mol 36.46 g = 0.411 mol HCl Since HCl is needed to react with Al in a 6:2 (i.e., 3:1) molar ratio, it seems pretty certain that HCl is the limiting reactant. To prove this, we can calculate the quantity of Al that would react with the given number of moles of HCl. 0.411 mol HCl 2 mol Al 6 mol HCl = 0.137 mol Al By this calculation we have shown that all the HCl present will be needed to react with only 0.137 mol Al (out of the 0.556 mol Al present). Therefore, HCl is the limiting reactant, and Al is present in excess. The calculation of the masses of products produced is based on the number of moles of the limiting reactant. 0.411 mol HCl 0.411 mol HCl b. 2 mol AlCl3 6 mol HCl 3 mol H 2 6 mol HCl 133.3 g 1 mol 2.016 g 1 mol = 18.3 g AlCl3 = 0.414 g H2 2NaOH(aq) + CO2(g) Na2CO3(aq) + H2O(l) molar masses: NaOH, 40.00 g; CO2, 44.01 g; Na2CO3, 105.99 g; H2O, 18.02 g 15.0 g NaOH 15.0 g CO2 1 mol 40.00 g 1 mol 44.01 g = 0.375 mol NaOH = 0.341 mol CO2 For the 0.375 mol NaOH, let's calculate if there is enough CO2 present to react: 0.375 mol NaOH World of Chemistry 1 mol CO 2 2 mol NaOH = 0.1875 mol CO2 (0.188 mol) Copyright Houghton Mifflin Company. All rights reserved. 86 Chapter 9 We have present more CO2 (0.341 mol) than is needed to react with the given quantity of NaOH. NaOH is therefore the limiting reactant, and CO2 is present in excess. The quantities of products resulting are based on the complete conversion of the limiting reactant (0.375 mol NaOH): 1 mol Na 2 CO3 0.375 mol NaOH 2 mol NaOH 1 mol H 2 O 0.375 mol NaOH c. 2 mol NaOH 105.99 g 1 mol 18.02 g 1 mol = 19.9 g Na2CO3 = 3.38 g H2O Pb(NO3)2(aq) + 2HCl(aq) PbCl2(s) + 2HNO3(aq) Molar masses: Pb(NO3)2, 331.2 g; HCl, 36.46 g; PbCl2, 278.1 g; HNO3, 63.02 g 15.0 g Pb(NO3)2 15.0 g HCl 1 mol 331.2 g 1 mol 36.46 g = 0.0453 mol Pb(NO3)2 = 0.411 mol HCl With such a large disparity between the numbers of moles of the reactants, it’s probably a sure bet that Pb(NO3)2 is the limiting reactant. To confirm this, we can calculate how many mol of HCl are needed to react with the given amount of Pb(NO3)2. 0.0453 mol Pb(NO3)2 2 mol HCl 1 mol Pb(NO3 ) 2 = 0.0906 mol HCl We have considerably more HCl present than is needed to react completely with the Pb(NO3)2. Therefore, Pb(NO3)2 is the limiting reactant, and HCl is present in excess. The quantities of products produced are based on the limiting reactant being completely consumed. 0.0453 mol Pb(NO3)2 0.0453 mol Pb(NO3)2 d. 1 mol PbCl2 1 mol Pb(NO 3 ) 2 2 mol HNO 3 1 mol Pb(NO 3 ) 2 278.1 g 1 mol 63.02 g 1 mol = 12.6 g PbCl2 = 5.71 g HNO3 2K(s) + I2(s) 2KI(s) Molar masses: K, 39.10 g; I2, 253.8 g; KI, 166.0 g 15.0 g K 15.0 g I2 World of Chemistry 1 mol 39.10 g 1 mol 253.8 g = 0.384 mol K = 0.0591 mol I2 Copyright Houghton Mifflin Company. All rights reserved. Chemical Quantities 87 Since, from the balanced chemical equation, we need twice as many moles of K as moles of I2, and since there is so little I2 present, it’s a safe bet that I2 is the limiting reactant. To confirm this, we can calculate how many moles of K are needed to react with the given amount of I2 present: 2 mol K 0.0591 mol I2 1 mol I 2 = 0.1182 mol K Clearly we have more potassium present than is needed to react with the small amount of I2 present. I2 is therefore the limiting reactant, and potassium is present in excess. The amount of product produced is calculated from the number of moles of the limiting reactant present: 2 mol KI 0.0591 mol I2 35. a. 1 mol I 2 166.0 g = 19.6 g KI 1 mol 2NH3(g) + 2Na(s) 2NaNH2(s) + H2(g) Molar masses: NH3, 17.03 g; Na, 22.99 g; NaNH2, 39.02 g 50.0 g NH3 50.0 g Na 1 mol = 2.94 mol NH3 17.03 g 1 mol 22.99 g = 2.17 mol Na Since the coefficients of NH3 and Na are the same in the balanced chemical equation for the reaction, the two reactants combine in a 1:1 molar ratio. Therefore, Na is the limiting reactant, which will control the amount of product produced. 2.17 mol Na b. 2 mol NaNH 2 2 mol Na 39.02 g 1 mol = 84.7 g NaNH2 BaCl2(aq) + Na2SO4(aq) BaSO4(s) + 2NaCl(aq) Molar masses: BaCl2, 208.2 g; Na2SO4, 142.1 g; BaSO4, 233.4 g 50.0 g BaCl2 1 mol 208.2 g 50.0 g Na2SO4 = 0.240 mol BaCl2 1 mol 142.1 g = 0.352 mol Na2SO4 Since the coefficients of BaCl2 and Na2SO4 are the same in the balanced chemical equation for the reaction, the reactant having the smaller number of moles present (BaCl2) must be the limiting reactant, which will control the amount of product produced. 0.240 mol BaCl2 World of Chemistry 1 mol BaSO 4 1 mol BaCl2 233.4 g 1 mol = 56.0 g BaSO4 Copyright Houghton Mifflin Company. All rights reserved. 88 Chapter 9 c. SO2(g) + 2NaOH(aq) Na2SO3(aq) + H2O(l) Molar masses: SO2, 64.07 g; NaOH, 40.00 g; Na2SO3, 126.1 g 50.0 g SO2 1 mol 64.07 g 50.0 g NaOH = 0.780 mol SO2 1 mol 40.00 g = 1.25 mol NaOH From the balanced chemical equation for the reaction, every time one mol of SO2 reacts, two mol of NaOH is needed. For 0.780 mol of SO2, 2(0.780 mol) = 1.56 mol of NaOH would be needed. We do not have sufficient NaOH to react with the SO2 present: therefore, NaOH is the limiting reactant, and controls the amount of product obtained. 1.25 mol NaOH d. 1 mol Na 2SO 3 2 mol NaOH 126.1 g 1 mol = 78.8 g Na2SO3 2Al(s) + 3H2SO4(l) Al2(SO4)3(s) + 3H2(g) Molar masses: Al, 26.98 g; H2SO4, 98.09 g; Al2(SO4)3, 342.2 g 50.0 g Al 1 mol 26.98 g 50.0 g H2SO4 = 1.85 mol Al 1 mol 98.09 g = 0.510 mol H2SO4 Since the amount of H2SO4 present is smaller than the amount of Al, let’s see if H2SO4 is the limiting reactant by calculating how much Al would react with the given amount of H2SO4. 0.510 mol H2SO4 2 mol Al 3 mol H 2SO 4 = 0.340 mol Al Since all the H2SO4 present would react with only a small portion of the Al present, H2SO4 is therefore the limiting reactant and will control the amount of product obtained. 0.510 mol H2SO4 36. 1 mol Al2 (SO 4 )3 3 mol H 2SO 4 342.2 g 1 mol = 58.2 g Al2(SO4)3 a. CO(g) + 2H2(g) CH3OH(l) CO is the limiting reactant; 11.4 mg CH3OH b. 2Al(s) + 3I2(s) 2AlI3(s) I2 is the limiting reactant; 10.7 mg AlI3 c. Ca(OH)2(aq) + 2HBr(aq) CaBr2(aq) + 2H2O(l) HBr is the limiting reactant; 12.4 mg CaBr2; 2.23 mg H2O World of Chemistry Copyright Houghton Mifflin Company. All rights reserved. Chemical Quantities 89 d. 2Cr(s) + 2H3PO4(aq) 2CrPO4(s) + 3H2(g) H3PO4 is the limiting reactant; 15.0 mg CrPO4; 0.309 mg H2 37. Cl2(g) + 2NaI(aq) 2NaCl(aq) + I2(s) Br2(l) + 2NaI(aq) 2NaBr(aq) + I2(s) molar masses: Cl2, 70.90 g; Br2, 159.8 g; I2, 253.8 g; NaI, 149.9 g 1 mol Cl2 5.00 g Cl2 70.90 g Cl2 25.0 g NaI = 0.0705 mol Cl2 1 mol NaI = 0.167 mol NaI 149.9 g NaI Since we would need 2(0.0705) = 0.141 mol of NaI for the Cl2 to react completely, and we have more than this amount of NaI, then Cl2 must be the limiting reactant. 0.0705 mol Cl2 0.0705 mol I2 1 mol I 2 1 mol Cl2 253.8 g I 2 1 mol I 2 1 mol Br2 5.00 g Br2 159.8 g Br2 25.0 g NaI = 0.0705 mol I2 = 17.9 g I2 = 0.0313 mol Br2 1 mol NaI 149.9 g NaI = 0.167 mol NaI Since we would need 2(0.0313) = 0.0626 mol of NaI for the Br2 to react completely, and we have more than this amount of NaI, then Br2 must be the limiting reactant. 0.0313 mol Br2 0.0313 mol I2 38. 1 mol I 2 1 mol Br2 253.8 g I 2 1 mol I 2 = 0.0313 mol I2 = 7.94 g I2 4Fe(s) + 3O2(g) 2Fe2O3(s) Molar masses: Fe, 55.85 g; Fe2O3, 159.7 g 1.25 g Fe 1 mol 55.85 g = 0.0224 mol Fe present Calculate how many mol of O2 are required to react with this amount of Fe. 0.0224 mol Fe World of Chemistry 3 mol O 2 4 mol Fe = 0.0168 mol O2 Copyright Houghton Mifflin Company. All rights reserved. 90 Chapter 9 Since we have more O2 than this, Fe must be the limiting reactant. 2 mol Fe 2 O3 0.0224 mol Fe 4 mol Fe 0.0112 mol Fe2O3 39. = 0.0112 mol Fe2O3 159.7 g Fe 2 O3 1 mol Fe 2 O3 = 1.79 g Fe2O3 Ca2+(aq) + Na2C2O4(aq) CaC2O4(s) + 2Na+(aq) molar masses: Ca2+, 40.08 g; Na2C2O4, 134.0 g 2+ 15 g Ca 1 mol Ca 2+ = 0.37 mol Ca2+ 40.08 g Ca 2+ 1 mol Na 2 C 2 O 4 15 g Na2C2O4 134.0 g Na 2 C 2 O 4 = 0.11 mol Na2C2O4 Since the balanced chemical equation tells us that one oxalate ion is needed to precipitate each calcium ion, from the number of moles calculated to be present, it should be clear that not nearly enough sodium oxalate ion has been added to precipitate all the calcium ion in the sample. 40. The actual yield for a reaction is the quantity of product actually isolated from the reaction vessel. The theoretical yield represents the mass of products that should be produced by the reaction if the limiting reactant is fully consumed. The percent yield represents what fraction of the amount of product that should have been collected was actually collected. 41. If the reaction is performed in a solvent, the product may have a substantial solubility in the solvent; the reaction may come to equilibrium before the full yield of product is achieved; loss of product may occur through operator error. 42. Percent yield = 43. S8(s) + 8Na2SO3(aq) + 40H2O(l) 8Na2S2O3•5H2O actual yield theoretical yield 100 = 1.23 g 1.44 g 100 = 85.4% molar masses: S8, 256.6 g; Na2SO3, 126.1 g; Na2S2O3•5H2O, 248.2 g 3.25 g S8 1 mol S8 256.6 g S8 13.1 g Na2SO3 = 0.01267 mol S8 1 mol Na 2SO3 126.1 g Na 2SO3 = 0.1039 mol Na2SO3 S8 is the limiting reactant. 0.01267 mol S8 World of Chemistry 8 mol Na 2S2 O3 • 5H 2 O 1 mol S8 = 0.1014 mol Na2S2O3•5H2O Copyright Houghton Mifflin Company. All rights reserved. Chemical Quantities 91 248.2 g Na 2S2 O3 • 5H 2 O 0.1014 mol Na2S2O3•5H2O actual yield Percent yield = 44. = 25.2 g Na2S2O3•5H2O 1 mol Na 2S2 O3 • 5H 2 O 100 = theoretical yield 5.26 g 100 = 20.9% 25.2 g 2LiOH(s) + CO2(g) Li2CO3(s) + H2O(g) molar masses: LiOH, 23.95 g; CO2, 44.01 g 1 mol LiOH 155 g LiOH 23.95 g LiOH 1 mol CO 2 2 mol LiOH 44.01 g CO 2 1 mol CO 2 = 142 g CO2 Since the cartridge has only absorbed 102 g CO2 out of a total capacity of 142 g CO2, the cartridge has absorbed 102 g 142 g 45. 100 = 71.8% of its capacity Xe(g) + 2F2(g) XeF4(s) molar masses: Xe, 131.3 g; F2, 38.00 g; XeF4, 207.3 g 130. g Xe 100. g F2 1 mol Xe = 0.9901 mol Xe 131.3 g Xe 1 mol F2 38.00 g F2 = 2.632 mol F2 Xe is the limiting reactant. 0.9901 mol Xe 1 mol XeF4 1 mol Xe 0.9901 mol XeF4 Percent yield = 46. = 0.9901 mol XeF4 207.3 g XeF4 1 mol XeF4 actual yield theoretical yield = 205 g XeF4 100 = 145 g 205 g 100 = 70.7% CaCO3(s) + 2HCl(g) CaCl2(s) + CO2(g) + H2O(g) molar masses: CaCO3, 100.1 g; HCl, 36.46 g; CaCl2, 111.0 g 155 g CaCO3 250. g HCl 1 mol CaCO3 100.1 g CaCO3 1 mol HCl 36.46 g HCl = 1.548 mol CaCO3 = 6.857 mol HCl CaCO3 is the limiting reactant. World of Chemistry Copyright Houghton Mifflin Company. All rights reserved. 92 Chapter 9 1.548 mol CaCO3 1.548 mol CaCl2 Percent yield = 47. 1 mol CaCl2 = 1.548 mol CaCl2 1 mol CaCO3 111.0 g CaCl2 = 172 g CaCl2 1 mol CaCl2 actual yield 100 = theoretical yield 142 g 172 g 100 = 82.6% NaCl(aq) + NH3(aq) + H2O(l) + CO2(s) NH4Cl(aq) + NaHCO3(s) molar masses: NH3, 17.03 g; CO2, 44.01 g; NaHCO3, 84.01 g 10.0 g NH3 15.0 g CO2 1 mol NH 3 17.03 g NH 3 1 mol CO 2 44.01 g CO 2 = 0.5872 mol NH3 = 0.3408 mol CO2 CO2 is the limiting reactant. 1 mol NaHCO3 0.3408 mol CO2 1 mol CO 2 0.3408 mol NaHCO3 48. = 0.3408 mol NaHCO3 84.01 g NaHCO3 1 mol NaHCO3 = 28.6 g NaHCO3 Fe(s) + S(s) FeS(s) molar masses: Fe, 55.85 g; S, 32.07 g; FeS, 87.92 g 5.25 g Fe 12.7 g S 1 mol Fe 55.85 g Fe 1 mol S 32.07 g S = 0.0940 mol Fe = 0.396 mol S Fe is the limiting reactant. 0.0940 mol Fe 49. 1 mol FeS 1 mol Fe 87.92 g FeS 1 mol FeS = 8.26 g FeS produced C6H12O6(s) + 6O2(g) 6CO2(g) + 6H2O(g) molar masses: glucose, 180.2 g; CO2, 44.01 g 1.00 g glucose 1 mol glucose 180.2 g glucose 5.549 10–3 mol glucose 3.33 10–2 mol CO2 World of Chemistry = 5.549 10–3 mol glucose 6 mol CO 2 1 mol glucose 44.01 g CO 2 1 mol CO 2 = 3.33 10–2 mol CO2 = 1.47 g CO2 Copyright Houghton Mifflin Company. All rights reserved. Chemical Quantities 50. 93 10.3 g Cl – mass of Cl present = 1.054 g sample 100.0 g sample = 0.1086 g Cl– molar masses: Cl–, 35.45 g; AgNO3, 169.9 g; AgCl, 143.4 g 0.1086 g Cl– 1 mol Cl = 3.063 10–3 mol Cl– 35.45 g Cl 3.063 10–3 mol Cl– 1 mol AgNO 3 1 mol Cl 3.063 10–3 mol AgNO3 3.063 10–3 mol Cl– 51. For O2: 52. a. 5 mol O 2 1 mol C3 H 8 = 0.520 g AgNO3 required = 3.063 10–3 mol AgCl 143.4 g AgCl 1 mol AgCl For CO2: = 0.439 g AgCl produced 3 mol CO 2 1 mol C3 H 8 For H2O: 4 mol H 2 O 1 mol C3 H 8 2H2O2(l) 2H2O(l) + O2(g) 0.50 mol H2O2 2 mol H 2 O 2 mol H 2 O 2 1 mol O 2 2 mol H 2 O 2 = 0.50 mol H2O = 0.25 mol O2 2KClO3(s) 2KCl(s) + 3O2(g) 0.50 mol KClO3 0.50 mol KClO3 c. 1 mol AgNO3 1 mol Cl- 0.50 mol H2O2 b. 169.9 g AgNO3 1 mol AgCl 3.063 10–3 mol AgCl = 3.063 10–3 mol AgNO3 2 mol KCl 2 mol KClO3 3 mol O 2 2 mol KClO3 = 0.50 mol KCl = 0.75 mol O2 2Al(s) + 6HCl(aq) 2AlCl3(aq) + 3H2(g) 0.50 mol Al 0.50 mol Al World of Chemistry 2 mol AlCl3 2 mol Al 3 mol H 2 2 mol Al = 0.50 mol AlCl3 = 0.75 mol H2 Copyright Houghton Mifflin Company. All rights reserved. 94 Chapter 9 d. C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l) 3 mol CO 2 0.50 mol C3H8 1 mol C3 H 8 4 mol H 2 O 0.50 mol C3H8 53. a. = 1.5 mol CO2 1 mol C3 H 8 = 2.0 mol H2O NH3(g) + HCl(g) NH4Cl(s) molar mass of NH3 = 17.0 g 1 mol NH 3 1.00 g NH3 17.0 g NH 3 0.0588 mol NH3 b. = 0.0588 mol NH3 1 mol NH 4 Cl 1 mol NH 3 = 0.0588 mol NH4Cl CaO(s) + CO2(g) CaCO3(s) molar mass CaO = 56.1 g 1.00 g CaO 1 mol CaO 56.1 g CaO 0.0178 mol CaO c. = 0.0178 mol CaO 1 mol CaCO3 1 mol CaO = 0.0178 mol CaCO3 4Na(s) + O2(g) 2Na2O(s) molar mass Na = 22.99 g 1.00 g Na 1 mol Na 0.0435 mol Na d. = 0.0435 mol Na 22.99 g Na 2 mol Na 2 O 4 mol Na = 0.0217 mol Na2O 2P(s) + 3Cl2(g) 2PCl3(l) molar mass P = 30.97 g 1.00 g P 1 mol P 30.97 g P 0.0323 mol P 54. = 0.0323 mol P 2 mol PCl3 2 mol P = 0.0323 mol PCl3 2Na2O2(s) + 2H2O(l) 4NaOH(aq) + O2(g) molar masses: Na2O2, 77.98 g; O2, 32.00 g 3.25 g Na2O2 World of Chemistry 1 mol Na 2 O 2 77.98 g Na 2 O 2 = 0.0417 mol Na2O2 Copyright Houghton Mifflin Company. All rights reserved. Chemical Quantities 95 0.0417 mol Na2O2 0.0209 mol O2 55. 1 mol O 2 2 mol Na 2 O 2 32.00 g O 2 = 0.0209 mol O2 = 0.669 g O2 1 mol O 2 2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(g) 150 g = 1.5 102 g molar masses: C2H2, 26.04 g; O2, 32.00 g 1 mol C 2 H 2 1.5 102 g C2H2 5.760 mol C2H2 14.40 mol O2 56. a. 26.04 g C 2 H 2 5 mol O 2 2 mol C 2 H 2 32.00 g O 2 1 mol O 2 = 5.760 mol C2H2 = 14.40 mol O2 = 4.6 102 g O2 C2H5OH(l) + 3O2(g) 2CO2(g) + 3H2O(l) molar masses: C2H5OH, 46.07 g; O2, 32.00 g; CO2, 44.01 g 1 mol C 2 H 5OH 25.0 g C2H5OH 25.0 g O2 46.07 g C 2 H 5OH 1 mol O 2 32.00 g O 2 = 0.5427 mol C2H5OH = 0.7813 mol O2 Since there is less C2H5OH present on a mole basis, see if this substance is the limiting reactant. 0.5427 mol C2H5OH 3 mol O 2 1 mol C 2 H 5OH = 1.6281 mol O2 From the above calculation, C2H5OH must not be the limiting reactant (even though there is a smaller number of moles of C2H5OH present) since more oxygen than is present would be required to react completely with the C2H5OH present. Oxygen is the limiting reactant. 0.7813 mol O2 2 mol CO 2 0.5209 mol CO2 b. 3 mol O 2 = 0.5209 mol CO2 44.01 g CO 2 1 mol CO 2 = 22.9 g CO2 N2(g) + O2(g) 2NO(g) molar masses: N2, 28.02 g; O2, 32.00 g; NO, 30.01 g 25.0 g N2 World of Chemistry 1 mol N 2 28.02 g N 2 = 0.8922 mol N2 Copyright Houghton Mifflin Company. All rights reserved. 96 Chapter 9 25.0 g O2 1 mol O 2 = 0.7813 mol O2 32.00 g O 2 Since the coefficients of N2 and O2 are the same in the balanced chemical equation for the reaction, an equal number of moles of each substance would be necessary for complete reaction. Since there is less O2 present on a mole basis, O2 must be the limiting reactant. 0.7813 mol O2 1.5626 mol NO c. 2 mol NO = 1.5626 mol NO 1 mol O 2 30.01 g NO 1 mol NO = 46.9 g NO 2NaClO2(aq) + Cl2(g) 2ClO2(g) + 2NaCl(aq) molar masses: NaClO2, 90.44 g; Cl2, 70.90 g; NaCl, 58.44 g 25.0 g NaClO2 25.0 g Cl2 1 mol NaClO 2 = 0.2764 mol NaClO2 90.44 g NaClO 2 1 mol Cl2 70.90 g Cl2 = 0.3526 mol Cl2 See if NaClO2 is the limiting reactant. 0.2764 mol NaClO2 1 mol Cl2 1 mol NaClO 2 = 0.1382 mol Cl2 Since 0.2764 mol of NaClO2 would require only 0.1382 mol Cl2 to react completely (and since we have more than this amount of Cl2), then NaClO2 must indeed be the limiting reactant. 0.2764 mol NaClO2 0.2764 mol NaCl d. 2 mol NaCl 2 mol NaClO 2 58.44 g NaCl 1 mol NaCl = 0.2764 mol NaCl = 16.2 g NaCl 3H2(g) + N2(g) 2NH3(g) molar masses: H2, 2.016 g; N2, 28.02 g; NH3, 17.03 g 25.0 g H2 25.0 g N2 1 mol H 2 2.016 g H 2 1 mol N 2 28.02 g N 2 = 12.40 mol H2 = 0.8922 mol N2 See if N2 is the limiting reactant. 0.8922 mol N2 World of Chemistry 3 mol H 2 1 mol N 2 = 2.677 mol H2 Copyright Houghton Mifflin Company. All rights reserved. Chemical Quantities 97 N2 is clearly the limiting reactant, since there is 12.40 mol H2 present (a large excess). 0.8922 mol N2 1.784 mol NH3 57. 2 mol NH 3 = 1.784 mol NH3 1 mol N 2 17.03 g NH 3 1 mol NH 3 = 30.4 g NH3 N2H4(l) + O2(g) N2(g) + 2H2O(g) molar masses: N2H4, 32.05 g; O2, 32.00 g; N2, 28.02 g; H2O, 18.02 g 20.0 g N2H4 20.0 g O2 1 mol N 2 H 2 = 0.624 mol N2H4 32.05 g N 2 H 2 1 mol O 2 32.00 g O 2 = 0.625 mol O2 The two reactants are present in nearly the required ratio for complete reaction (due to the 1:1 stoichiometry of the reaction and the very similar molar masses of the substances). We will consider N2H4 as the limiting reactant in the following calculations. 1 mol N 2 0.624 mol N2H4 0.624 mol N2 1 mol N 2 H 4 28.02 g N 2 1 mol N 2 1.248 mol H2O = 17.5 g N2 2 mol H 2 O 0.624 mol N2H4 = 0.624 mol N2 1 mol N 2 H 4 18.02 g H 2 O 1 mol H 2 O = 1.248 mol H2O = 1.25 mol H2O = 22.5 g H2O 40 g actual 58. 12.5 g theoretical 59. We are concerned with the balanced equation: 100 g theoretical = 5.0 g X4 + 12HCl 4XCl3 + 6H2 We are given the mass of X4 (248 g). If we can determine the number of moles of X4, we can determine its molar mass, and therefore its identity. We can do this from the mass of H2. 24.0 g H2 1 mol H 2 2.016 g H 2 1 mol X 4 6 mol H 2 = 1.98 mol X4 Thus, 248 g X 4 1.98 mol X 4 World of Chemistry = 125 g/mol (molar mass of X4) Copyright Houghton Mifflin Company. All rights reserved. 98 Chapter 9 Element X has an atomic mass of about 125 = 31.3 g/mol. 4 This is closest to phosphorus (30.97 g/mol). 60. We know the following: x + y xy 5 g 15 g x + 3z xz3 3 g 18 g and Thus, the mole ratio between x and y is 1:1 and the mole ratio between x and z is 1:3, respectively. So, the relative masses of x:y are 1:3, and the relative masses of x:z are 3:6 or 1:2 (if x = 3 g, 3z = 18 g or z = 6 g). Thus, x:y:z = 1:3:2. If y = 60 g/mol, x = 20 g/mol, and z = 40 g/mol. 61. This is a “double” limiting reactant problem. First, determine the maximum yield of P4O10 from the equation: P4 + 5O2 P4O10 20.0 g P4 30.0 g O2 1 mol P4 123.88 g P4 1 mol O 2 32.00 g O 2 1 mol P4 O10 1 mol P4 1 mol P4 O10 5 mol O 2 283.88 g P4 O10 1 mol P4 O10 283.88 g P4 O10 1 mol P4 O10 = 45.8 g P4O10 = 53.2 g P4O10 Thus, the maximum yield of P4O10 is 45.8 g. Determine the yield of H3PO4 from the equation: P4O10 + 6H2O 4H3PO4 45.8 g P4O10 15.0 g H2O 1 mol P4 O10 283.88 g P4 O10 1 mol H 2 O 18.016 g H 2 O 4 mol H 3 PO 4 1 mol P4 O10 4 mol H 3 PO 4 6 mol H 2 O 97.994 g H 3 PO 4 1 mol H 3 PO 4 97.994 g H 3 PO 4 1 mol H 3 PO 4 = 63.2 g H3PO4 = 54.4 g H3PO4 Thus, the maximum yield of H3PO4 is 54.4 g. 62. a. Neither is limiting. 1 mol N2 2 mol NH 3 1 mol N 2 = 2 moles NH3 3 mol H2 2 mol NH 3 = 2 moles NH3 3 mol H 2 b. H2 is limiting. 3 mol H2 2 mol NH 3 = 2 moles NH3 3 mol H 2 c. H2 is limiting. 3 mol H2 2 mol NH 3 = 2 moles NH3 3 mol H 2 d. So, choice d (each would produce the same amount of product) is the correct answer. World of Chemistry Copyright Houghton Mifflin Company. All rights reserved. Chemical Quantities 99 63. Choice “c” is correct. Since the molar mass of O2 is larger than the molar mass of C2H6, an equal mass of each would result in a fewer number of moles of O2 than C2H6. If this were the case, O2 must be limiting since it is needed in a greater amount (a 7:2 mol ratio). 64. 50.0 g CaO 1 mol CaO 56.08 g CaO 5 mol C = 2.23 mol C required 2 mol CaO 50.0 g C 1 mol C = 4.16 mol C 12.01 g C Leftover C = 4.16 mol – 2.23 mol = 1.93 mol C 12.01 g C = 23.2 g C left over 1 mol C (CaO is limiting.) 65. 13.97 g KClO3 Percent yield = 1 mol KClO3 74.55 g KCl = 8.498 g KCl 2 mol KCl 122.55 g KClO3 2 mol KClO3 1 mol KCl actual yield 6.23 g 100% = 100% = 73.3% yield theoretical yield 8.498 g 66. False. Amounts used can vary. However, according to the equation, for every 28.02 g of N2 (2 mol) we need 6.048 g of H2 (3 mol). So, for example, if we have 28.02 g of N2 and 5.5 g of H2, the H2 is limiting. If we have 28.02 g of N2 and 6.5 g of H2, the N2 is limiting. 67. a. b. As we add more sodium, we get more product because sodium is the limiting reactant. However, eventually sodium is in excess (and chlorine is limiting), so the amount of product does not increase with an increase in sodium. 58.44 g NaCl 20.0 g Na 1 mol Na 2 mol NaCl = 50.8 g NaCl 22.99 g Na 2 mol Na 1 mol NaCl c. If we look at the graph, we can see that 40.0 g of sodium reacts in a stoichiometric ratio with chlorine (that is, there is no limiting reactant). Because of this, we can use 40.0 g of sodium to determine the amount of Cl2. Since the problem states that the mass of Cl2 is the same in all containers, each container will have this amount of Cl2. 1 mol Cl2 70.9 g Cl2 40.0 g Na 1 mol Na = 61.7 g Cl2 22.99 g Na d. 1 mol Cl2 We should add “since the amount of product is equal when 40.0 g or 50.0 g of sodium is used, 50.0 g of sodium must be an excess amount.” Since we know from part c how much Cl2 is in the container, we use the amount of Cl2 (the limiting reactant) to determine the amount of product. 61.7 g Cl2 e. 2 mol Na 1 mol Cl2 58.44 g NaCl = 101.7 g NaCl = 102 g NaCl 2 mol NaCl 70.9 g Cl2 1 mol Cl2 1 mol NaCl In part b, sodium is limiting, and 30.85 g Cl2 is left over (half of the Cl2 is left over). In part d, Cl2 is limiting and 10.0 g Na is left over (50.0 g – 40.0 g). World of Chemistry Copyright Houghton Mifflin Company. All rights reserved.