c. PCl 3 (l) + 3H 2 O(l) H 3 PO 3 (l) + 3HCl(g)

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1 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 O 2 (g) are different, 1 g of O 2 could not represent the same number of moles as 1 g of C. 3. a. 2NO(g) + O 2 (g) 2NO 2 (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. 2AgC 2 H 3 O 2 (aq) + CuSO 4 (aq) Ag 2 SO 4 (s) + Cu(C 2 H 3 O 2 ) 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. PCl 3 (l) + 3H 2 O(l) H 3 PO 3 (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. C 2 H 6 (g) + Cl 2 (g) C 2 H 5 Cl(g) + HCl(g) One molecule of ethane (C 2 H 6 ) reacts with one molecule of chlorine, producing one molecule of chloroethane (C 2 H 5 Cl) 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. 3MnO 2 (s) + 4Al(s) 3Mn(s) + 2Al 2 O 3 (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. 74

2 Chemical Quantities 75 b. B 2 O 3 (s) + 3CaF 2 (s) 2BF 3 (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. 3NO 2 (g) + H 2 O(l) 2HNO 3 (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. C 6 H 6 (g) + 3H 2 (g) C 6 H 12 (g) One molecule of C 6 H 6 (which is named benzene) reacts with three molecules of hydrogen, producing just one molecule of C 6 H 12 (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 CH 4 to the number of moles of oxygen needed for reaction, the correct mole ratio is 2 mol O 2 CH 4 For converting from a given number of moles of CH 4 to the number of moles of product produced, the ratios are CO 2 CH 4 and 2 mol H 2 O CH Ag(s) + H 2 S(g) Ag 2 S(s) + H 2 (g) Ag 2 S 2 mol Ag and H 2 2 mol Ag 8. a. 2FeO(s) + C(s) 2Fe(l) + CO 2 (g) mol FeO 2 mol Fe 2 mol FeO = mol Fe mol FeO CO 2 2 mol FeO = mol CO 2

3 76 Chapter 9 b. Cl 2 (g) + 2KI(aq) 2KCl(aq) + I 2 (s) mol KI 2 mol KCl 2 mol KI = mol KCl mol KI I 2 2 mol KI = mol I 2 c. Na 2 B 4 O 7 (s) + H 2 SO 4 (aq) + 5H 2 O(l) 4H 3 BO 3 (s) + Na 2 SO 4 (aq) mol Na 2 B 4 O 7 4 mol H 3 BO 3 Na 2 B 4 O 7 = mol H 3 BO mol Na 2 B 4 O 7 Na 2 SO 4 Na 2 B 4 O 7 = mol Na 2 SO 4 d. CaC 2 (s) + 2H 2 O(l) Ca(OH) 2 (s) + C 2 H 2 (g) mol CaC 2 Ca(OH) 2 CaC 2 = mol Ca(OH) mol CaC 2 C 2 H 2 CaC 2 = mol C 2 H 2 9. a. NH 3 (g) + HCl(g) NH 4 Cl(s) molar mass of NH 4 Cl, g 0.50 mol NH 3 NH 4 Cl NH 3 = 0.50 mol NH 4 Cl 0.50 mol NH 4 Cl g NH 4 Cl NH 4 Cl = 27 g NH 4 Cl b. CH 4 (g) + 4S(g) CS 2 (l) + 2H 2 S(g) molar masses: CS 2, g; H 2 S, g 0.50 mol S CS 2 4 mol S = mol CS 2 (= 0.13 mol CS 2 ) mol CS g CS 2 CS 2 = 9.5 g CS mol S 2 mol H 2 S 4 mol S = 0.25 mol H 2 S 0.25 mol H 2 S g H 2 S H 2 S = 8.5 g H 2S c. PCl 3 (l) + 3H 2 O(l) H 3 PO 3 (aq) + 3HCl(aq)

4 Chemical Quantities 77 molar masses: H 3 PO 3, g; HCl, g 0.50 mol PCl 3 H 3 PO 3 PCl 3 = 0.50 mol H 3 PO mol H 3 PO g H 3 PO 3 H 3 PO 3 = 41 g H 3 PO mol PCl 3 3 mol HCl PCl 3 = 1.5 mol HCl 1.50 mol HCl g HCl HCl = 54.7 = 55 g HCl d. NaOH(s) + CO 2 (g) NaHCO 3 (s) molar mass of NaHCO 3, g 0.50 mol NaOH NaHCO 3 NaOH = 0.50 mol NaHCO mol NaHCO g NaHCO 3 NaHCO 3 = 42 g NaHCO Before doing the calculations, the equations must be balanced. a. 4KO 2 (s) + 2H 2 O(l) 3O 2 (g) + 4KOH(s) mol KOH 3 mol O 2 4 mol KOH = mol O 2 b. SeO 2 (g) + 2H 2 Se(g) 3Se(s) + 2H 2 O(g) mol H 2 O 3 mol Se 2 mol H 2 O = mol Se c. 2CH 3 CH 2 OH(l) + O 2 (g) 2CH 3 CHO(aq) + 2H 2 O(l) mol H 2 O 2 mol CH 3 CHO 2 mol H 2 O = mol CH 3 CHO d. Fe 2 O 3 (s) + 2Al(s) 2Fe(l) + Al 2 O 3 (s) mol Al 2 O the molar mass of the substance 2 mol Fe Al 2 O 3 = 1.25 mol Fe 12. Stoichiometry is the process of using a chemical equation to calculate the relative masses of reactants and products involved in a reaction.

5 78 Chapter a. molar mass of Ag = g g Ag g = mol Ag b. molar mass of (NH 4 ) 2 S = g 1 g 45.2 mg (NH 4 ) 2 S 1000 mg g = mol (NH 4 ) 2 S c. molar mass of uranium = g 61.7 g U 1 g 10 6 g d. molar mass of SO 2 = g 5.23 kg SO g 1 kg e. molar mass of Fe(NO 3 ) 3 = g g = mol U g = 81.6 mol SO g Fe(NO 3 ) g = 1.12 mol Fe(NO 3) 3 f. molar mass of FeSO 4 = g 12.7 mg FeSO 4 1 g 1000 mg g = mol FeSO 4 g. molar mass of LiOH = g g LiOH = = 289 mol LiOH g 14. a. molar mass of CaCO 3 = g mol CaCO g = g CaCO 3 b. molar mass of He = g 2.75 mol He g = 11.0 g He c. molar mass of O 2 = g mol O g = g O 2 d. molar mass of CO 2 = g 7.21 millimol = CO g = g CO 2

6 Chemical Quantities 79 e. molar mass of FeS = g mol FeS g = 73.4 g FeS f. molar mass of KOH = g 4.0 KOH g = 225 g KOH g. molar mass of H 2 = g mol H g = g H Before any calculations are done, the equations must be balanced. a. Mg(s) + CuCl 2 (aq) MgCl 2 (aq) + Cu(s) molar mass of Mg = g 25.0 g Mg = 1.03 mol Mg g 1.03 mol Mg CuCl 2 Mg = 1.03 mol CuCl 2 b. 2AgNO 3 (aq) + NiCl 2 (aq) 2AgCl(s) + Ni(NO 3 ) 2 (aq) molar mass of AgNO 3 = g 25.0 g AgNO g = mol AgNO mol AgNO 3 NiCl 2 2 mol AgNO 3 = mol NiCl 2 c. NaHSO 3 (aq) + NaOH(aq) Na 2 SO 3 (aq) + H 2 O(l) molar mass of NaHSO 3 = g 25.0 g NaHSO g = mol NaHSO mol NaHSO 3 NaOH NaHSO 3 = mol NaOH d. KHCO 3 (aq) + HCl(aq) KCl(aq) + H 2 O(l) + CO 2 (g) molar mass of KHCO 3 = g 25.0 g KHCO g = mol KHCO mol KHCO 3 HCl KHCO 3 = mol HCl

7 80 Chapter 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. FeSO 4 (aq) + K 2 CO 3 (aq) FeCO 3 (s) + K 2 SO 4 (aq) millimolar masses: FeSO 4, mg; FeCO 3, mg; K 2 SO 4, mg 10.0 mg FeSO 4 1 mmol FeSO mg FeSO 4 = mmol FeSO mmol FeSO 4 1 mmol FeCO 3 1 mmol FeSO mg FeCO 3 1 mmol FeCO 3 = 7.63 mg FeCO mmol FeSO 4 1 mmol K 2 SO 4 1 mmol FeSO mg K 2 SO 4 1 mmol K 2 SO 4 = 11.5 mg K 2 SO 4 b. 4Cr(s) + 3SnCl 4 (l) 4CrCl 3 (s) + 3Sn(s) millimolar masses: Cr, mg; CrCl 3, mg; Sn, mg 10.0 mg Cr 1 mmol Cr mg Cr = mmol Cr mmol Cr 4 mmol CrCl 3 4 mmol Cr mg CrCl 3 1 mmol CrCl 3 = 30.4 mg CrCl mmol Cr 3 mmol Sn 4 mmol Cr mg Sn 1 mmol Sn = 17.1 mg Sn c. 16Fe(s) + 3S 8 (s) 8Fe 2 S 3 (s) millimolar masses: S 8, mg; Fe 2 S 3, mg 10.0 mg S 8 1 mmol S mg S 8 = mmol S mmol S 8 8 mmol Fe 2 S 3 3 mmol S mg Fe 2 S 3 1 mmol Fe 2 S 3 = 21.6 mg Fe 2 S 3 d. 3Ag(s) + 4HNO 3 (aq) 3AgNO 3 (aq) + 2H 2 O(l) + NO(g) millimolar masses: HNO 3, 63.0 mg; AgNO 3, mg H 2 O, 18.0 mg; NO, 30.0 mg 10.0 mg HNO 3 1 mmol HNO mg HNO 3 = mmol HNO mmol HNO 3 3 mmol AgNO 3 4 mmol HNO mg AgNO 3 1 mmol AgNO 3 = 20.3 mg AgNO 3

8 Chemical Quantities mmol HNO 3 2 mmol H 2 O 4 mmol HNO mg H 2 O 1 mmol H 2 O = 1.43 mg H 2O mmol HNO 3 1 mmol NO 4 mmol HNO mg NO 1 mmol NO = 1.19 mg NO 17. 2H 2 (g) + O 2 (g) 2H 2 O(g) molar masses: H 2, g; H 2 O, g 56.0 g H 2 H g H 2 = mol H mol H 2 2 mol H 2 O 2 mol H 2 = mol H 2 O mol H 2 O g H 2 O H 2 O = 500. g H 2O 18. 2H 2 (g) + O 2 (g) 2H 2 O(g) molar masses of O 2 = g mol H 2 O 2 2 mol H 2 = = mol O mol O g O 2 O 2 = g O molar masses: C, g; CO, g; CO 2, g 5.00 g C C = mol C g C carbon dioxide: C(s) + O 2 (g) CO 2 (g) mol C CO 2 C = mol CO mol CO g CO 2 CO 2 = 18.3 g CO 2 carbon monoxide: 2C(s) + O 2 (g) 2CO(g) mol C 2 mol CO 2 mol C = mol CO mol CO g CO CO = 11.7 g CO

9 82 Chapter Fe(s) + 3Cl 2 (g) 2FeCl 3 (s) millimolar masses: Fe, mg; FeCl 3, mg 15.5 mg Fe 1 mmol Fe mg Fe = mmol Fe mmol Fe 2 mmol FeCl 3 2 mmol Fe = mmol FeCl mmol FeCl mg FeCl 3 1 mmol FeCl 3 = 45.0 mg FeCl H 2 O 2 (aq) 2H 2 O(l) + O 2 (g) molar masses: H 2 O 2, g; O 2, g g H 2 O 2 H 2 O g H 2 O 2 = mol H 2 O mol H 2 O 2 O 2 2 mol H 2 O 2 = mol O mol O g O 2 O 2 = g O Cu(s) + S(s) CuS(s) molar masses: Cu, g; S, g 1.25 g Cu g = mol Cu mol Cu S Cu = mol S mol S g = g S 23. 2NH 4 NO 3 (s) 2N 2 (g) + O 2 (g) + 4H 2 O(g) molar masses: NH 4 NO 3, g; N 2, g; O 2, g; H 2 O, g 1.25 g NH 4 NO 3 NH 4 NO g NH 4 NO 3 = mol NH 4 NO mol NH 4 NO 3 2 mol N 2 2 mol NH 4 NO 3 = mol N mol N g N 2 N 2 = g N 2

10 Chemical Quantities mol NH 4 NO 3 O 2 2 mol NH 4 NO 3 = mol O mol O g O 2 O 2 = g O mol NH 4 NO 3 4 mol H 2 O 2 mol NH 4 NO 3 = mol H 2 O mol H 2 O g H 2 O H 2 O = g H 2O As a check, note that g g g = g = 1.25 g Mg(s) + O 2 (g) 2MgO(s) molar masses: Mg, g; MgO, g 1.25 g Mg g = mol Mg mol Mg 2 mol MgO 2 mol Mg = mol MgO mol MgO g = 2.07 g MgO 25. From the balanced equation: Cl 2 + 2KI I 2 + 2KCl We can calculate the following: g Cl 2 Cl g Cl 2 I 2 Cl g I 2 I 2 = g I From the balanced equation: Cl 2 + F 2 2ClF We can calculate the following: g ClF 27. Start by balancing the equations: C 3 H 8 + 5O 2 3CO 2 + 4H 2 O 2C 4 H O 2 8CO H 2 O ClF g ClF Cl 2 2 mol ClF 70.9 g Cl 2 Cl 2 = g Cl 2 If we had, for example, g of C 3 H 8 and g of C 4 H 10, we can calculate the mass of oxygen required for each reaction and compare the results.

11 84 Chapter g C 3 H 8 C 3 H g C 3 H 8 5 mol O 2 C 3 H g O 2 O 2 = g O g C 4 H 10 C 4 H g C 4 H mol O 2 2 mol C 4 H g O 2 O 2 = g O 2 Therefore, more O 2 would be required for the combustion of C 3 H 8. The same result is obtained with masses other than g. 28. Start with the balanced equations: CH 4 + 2O 2 CO 2 + 2H 2 O 4NH 3 + 5O 2 4NO + 6H 2 O The amount of water produced from 1.00 g CH 4 is: 1.00 g CH 4 CH g CH 4 2 mol H 2 O CH 4 = mol H 2 O The mass of NH 3 needed to produce mol H 2 O is: mol H 2 O 4 mol NH 3 6 mol H 2 O g NH 3 NH 3 = 1.42 g NH 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 N 2 and 6 molecules H 2, and these react according to the balanced equation: N 2 + 3H 2 2NH 3 We can determine amounts of product and leftover reactant from this information given. N 2 + 3H 2 2NH 3 start react end 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 N 2 and 4 molecules NH 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.

12 Chemical Quantities 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) 2AlCl 3 (aq) + 3H 2 (g) Molar masses: Al, g; HCl, g; AlCl 3, g; H 2, g 15.0 g Al g = mol Al 15.0 g HCl g = 0.41 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. 2 mol Al 0.41 HCl = mol Al 6 mol HCl By this calculation we have shown that all the HCl present will be needed to react with only mol Al (out of the 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 HCl 2 mol AlCl 3 6 mol HCl g = 18.3 g AlCl HCl 3 mol H 2 6 mol HCl g = g H 2 b. 2NaOH(aq) + CO 2 (g) Na 2 CO 3 (aq) + H 2 O(l) molar masses: NaOH, g; CO 2, g; Na 2 CO 3, g; H 2 O, g 15.0 g NaOH g = mol NaOH 15.0 g CO g = 0.34 CO 2 For the mol NaOH, let's calculate if there is enough CO 2 present to react: mol NaOH CO 2 2 mol NaOH = mol CO 2 (0.188 mol)

13 86 Chapter 9 We have present more CO 2 (0.34) than is needed to react with the given quantity of NaOH. NaOH is therefore the limiting reactant, and CO 2 is present in excess. The quantities of products resulting are based on the complete conversion of the limiting reactant (0.375 mol NaOH): mol NaOH Na 2 CO 3 2 mol NaOH g = 19.9 g Na 2 CO mol NaOH H 2 O 2 mol NaOH g = 3.38 g H 2 O c. Pb(NO 3 ) 2 (aq) + 2HCl(aq) PbCl 2 (s) + 2HNO 3 (aq) Molar masses: Pb(NO 3 ) 2, g; HCl, g; PbCl 2, g; HNO 3, g 15.0 g Pb(NO 3 ) g = mol Pb(NO 3) g HCl g = 0.41 HCl With such a large disparity between the numbers of moles of the reactants, it s probably a sure bet that Pb(NO 3 ) 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(NO 3 ) 2. 2 mol HCl mol Pb(NO 3 ) 2 = mol HCl Pb(NO 3 ) 2 We have considerably more HCl present than is needed to react completely with the Pb(NO 3 ) 2. Therefore, Pb(NO 3 ) 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 mol Pb(NO 3 ) 2 PbCl 2 Pb(NO 3 ) g mol Pb(NO 3 ) 2 2 mol HNO 3 Pb(NO 3 ) g = 12.6 g PbCl 2 = 5.71 g HNO 3 d. 2K(s) + I 2 (s) 2KI(s) Molar masses: K, g; I 2, g; KI, g 15.0 g K = mol K g 15.0 g I g = I 2

14 Chemical Quantities 87 Since, from the balanced chemical equation, we need twice as many moles of K as moles of I 2, and since there is so little I 2 present, it s a safe bet that I 2 is the limiting reactant. To confirm this, we can calculate how many moles of K are needed to react with the given amount of I 2 present: I 2 2 mol K = mol K I 2 Clearly we have more potassium present than is needed to react with the small amount of I 2 present. I 2 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: I 2 2 mol KI g I 2 = 19.6 g KI 35. a. 2NH 3 (g) + 2Na(s) 2NaNH 2 (s) + H 2 (g) Molar masses: NH 3, g; Na, g; NaNH 2, g 50.0 g NH g = 2.94 mol NH g Na g = 2.17 mol Na Since the coefficients of NH 3 and Na are the same in the balanced chemical equation for the reaction, the two reactants combine in a 1:ar ratio. Therefore, Na is the limiting reactant, which will control the amount of product produced mol Na 2 mol NaNH 2 2 mol Na g = 84.7 g NaNH 2 b. BaCl 2 (aq) + Na 2 SO 4 (aq) BaSO 4 (s) + 2NaCl(aq) Molar masses: BaCl 2, g; Na 2 SO 4, g; BaSO 4, g 50.0 g BaCl g = mol BaCl g Na 2 SO g = mol Na 2SO 4 Since the coefficients of BaCl 2 and Na 2 SO 4 are the same in the balanced chemical equation for the reaction, the reactant having the smaller number of moles present (BaCl 2 ) must be the limiting reactant, which will control the amount of product produced mol BaCl 2 BaSO g = 56.0 g BaSO 4 BaCl 2

15 88 Chapter 9 c. SO 2 (g) + 2NaOH(aq) Na 2 SO 3 (aq) + H 2 O(l) Molar masses: SO 2, g; NaOH, g; Na 2 SO 3, g 50.0 g SO g = mol SO g NaOH g = 1.25 mol NaOH From the balanced chemical equation for the reaction, every time one mol of SO 2 reacts, two mol of NaOH is needed. For mol of SO 2, 2(0.780 mol) = 1.56 mol of NaOH would be needed. We do not have sufficient NaOH to react with the SO 2 present: therefore, NaOH is the limiting reactant, and controls the amount of product obtained mol NaOH Na 2 SO 3 2 mol NaOH g d. 2Al(s) + 3H 2 SO 4 (l) Al 2 (SO 4 ) 3 (s) + 3H 2 (g) = 78.8 g Na 2 SO 3 Molar masses: Al, g; H 2 SO 4, g; Al 2 (SO 4 ) 3, g 50.0 g Al = 1.85 mol Al g 50.0 g H 2 SO g = mol H 2SO 4 Since the amount of H 2 SO 4 present is smaller than the amount of Al, let s see if H 2 SO 4 is the limiting reactant by calculating how much Al would react with the given amount of H 2 SO 4. 2 mol Al mol H 2 SO 4 = mol Al 3 mol H 2 SO 4 Since all the H 2 SO 4 present would react with only a small portion of the Al present, H 2 SO 4 is therefore the limiting reactant and will control the amount of product obtained mol H 2 SO 4 Al (SO ) g 3 mol H 2 SO 4 = 58.2 g Al 2 (SO 4 ) a. CO(g) + 2H 2 (g) CH 3 OH(l) CO is the limiting reactant; 11.4 mg CH 3 OH b. 2Al(s) + 3I 2 (s) 2AlI 3 (s) I 2 is the limiting reactant; 10.7 mg AlI 3 c. Ca(OH) 2 (aq) + 2HBr(aq) CaBr 2 (aq) + 2H 2 O(l) HBr is the limiting reactant; 12.4 mg CaBr 2 ; 2.23 mg H 2 O

16 Chemical Quantities 89 d. 2Cr(s) + 2H 3 PO 4 (aq) 2CrPO 4 (s) + 3H 2 (g) H 3 PO 4 is the limiting reactant; 15.0 mg CrPO 4 ; mg H Cl 2 (g) + 2NaI(aq) 2NaCl(aq) + I 2 (s) Br 2 (l) + 2NaI(aq) 2NaBr(aq) + I 2 (s) molar masses: Cl 2, g; Br 2, g; I 2, g; NaI, g 5.00 g Cl 2 Cl g Cl 2 = mol Cl g NaI NaI g NaI = mol NaI Since we would need 2(0.0705) = 0.14 of NaI for the Cl 2 to react completely, and we have more than this amount of NaI, then Cl 2 must be the limiting reactant mol Cl 2 I 2 Cl 2 = mol I mol I g I 2 I 2 = 17.9 g I g Br 2 Br g Br 2 = mol Br g NaI NaI g NaI = mol NaI Since we would need 2(0.0313) = mol of NaI for the Br 2 to react completely, and we have more than this amount of NaI, then Br 2 must be the limiting reactant mol Br 2 I 2 Br 2 = mol I mol I g I 2 I 2 = 7.94 g I Fe(s) + 3O 2 (g) 2Fe 2 O 3 (s) Molar masses: Fe, g; Fe 2 O 3, g 1.25 g Fe = mol Fe present g Calculate how many mol of O 2 are required to react with this amount of Fe mol Fe 3 mol O 2 4 mol Fe = mol O 2

17 90 Chapter 9 Since we have more O 2 than this, Fe must be the limiting reactant mol Fe 2 mol Fe 2 O 3 4 mol Fe = mol Fe 2 O mol Fe 2 O g Fe 2 O 3 Fe 2 O 3 = 1.79 g Fe 2 O Ca 2+ (aq) + Na 2 C 2 O 4 (aq) CaC 2 O 4 (s) + 2Na + (aq) molar masses: Ca 2+, g; Na 2 C 2 O 4, g 15 g Ca Ca = 0.37 mol Ca g Ca g Na 2 C 2 O 4 Na 2 C 2 O g Na 2 C 2 O 4 = 0.1 Na 2 C 2 O 4 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 = actual yield theoretical yield 100 = 1.23 g 1.44 g 100 = 85.4% 43. S 8 (s) + 8Na 2 SO 3 (aq) + 40H 2 O(l) 8Na 2 S 2 O 3 5H 2 O molar masses: S 8, g; Na 2 SO 3, g; Na 2 S 2 O 3 5H 2 O, g 3.25 g S 8 S g S 8 = mol S g Na 2 SO 3 Na 2 SO g Na 2 SO 3 = mol Na 2 SO 3 S 8 is the limiting reactant mol S 8 8 mol Na 2 S 2 O 3 5H 2 O S 8 = mol Na 2 S 2 O 3 5H 2 O

18 Chemical Quantities mol Na 2 S 2 O 3 5H 2 O g Na 2 S 2 O 3 5H 2 O Na 2 S 2 O 3 5H 2 O = 25.2 g Na 2S 2 O 3 5H 2 O Percent yield = actual yield theoretical yield 100 = 5.26 g 25.2 g 100 = 20.9% 44. 2LiOH(s) + CO 2 (g) Li 2 CO 3 (s) + H 2 O(g) molar masses: LiOH, g; CO 2, g 155 g LiOH LiOH g LiOH CO 2 2 mol LiOH g CO 2 CO 2 = 142 g CO 2 Since the cartridge has only absorbed 102 g CO 2 out of a total capacity of 142 g CO 2, the cartridge has absorbed 102 g 142 g 100 = 71.8% of its capacity 45. Xe(g) + 2F 2 (g) XeF 4 (s) molar masses: Xe, g; F 2, g; XeF 4, g 130. g Xe Xe g Xe = Xe 100. g F 2 F g F 2 = mol F 2 Xe is the limiting reactant Xe XeF 4 Xe = XeF XeF g XeF 4 XeF 4 = 205 g XeF 4 Percent yield = actual yield theoretical yield 100 = 145 g 205 g 100 = 70.7% 46. CaCO 3 (s) + 2HCl(g) CaCl 2 (s) + CO 2 (g) + H 2 O(g) molar masses: CaCO 3, g; HCl, g; CaCl 2, g 155 g CaCO 3 CaCO g CaCO 3 = mol CaCO g HCl HCl g HCl = mol HCl CaCO 3 is the limiting reactant.

19 92 Chapter mol CaCO 3 CaCl 2 CaCO 3 = mol CaCl mol CaCl g CaCl 2 CaCl 2 = 172 g CaCl 2 Percent yield = actual yield theoretical yield 100 = 142 g 172 g 100 = 82.6% 47. NaCl(aq) + NH 3 (aq) + H 2 O(l) + CO 2 (s) NH 4 Cl(aq) + NaHCO 3 (s) molar masses: NH 3, g; CO 2, g; NaHCO 3, g 10.0 g NH 3 NH g NH 3 = mol NH g CO 2 CO 2 = mol CO g CO 2 CO 2 is the limiting reactant mol CO 2 NaHCO 3 CO 2 = mol NaHCO mol NaHCO g NaHCO 3 NaHCO 3 = 28.6 g NaHCO Fe(s) + S(s) FeS(s) molar masses: Fe, g; S, g; FeS, g Fe 5.25 g Fe = mol Fe g Fe 12.7 g S S = mol S g S Fe is the limiting reactant. FeS g FeS mol Fe Fe FeS = 8.26 g FeS produced 49. C 6 H 12 O 6 (s) + 6O 2 (g) 6CO 2 (g) + 6H 2 O(g) molar masses: glucose, g; CO 2, g 1.00 g glucose glucose g glucose = mol glucose mol glucose 6 mol CO 2 glucose = mol CO mol CO g CO 2 CO 2 = 1.47 g CO 2

20 Chemical Quantities mass of Cl present = g sample 10.3 g Cl g sample = g Cl molar masses: Cl, g; AgNO 3, g; AgCl, g g Cl Cl g Cl = mol Cl mol Cl AgNO 3 Cl = mol AgNO mol AgNO g AgNO 3 AgNO 3 = g AgNO 3 required mol Cl AgCl Cl - = mol AgCl mol AgCl g AgCl AgCl = g AgCl produced 51. For O 2 : 5 mol O 2 C 3 H 8 For CO 2 : 3 mol CO 2 C 3 H 8 For H 2 O: 4 mol H 2 O C 3 H a. 2H 2 O 2 (l) 2H 2 O(l) + O 2 (g) 0.50 mol H 2 O 2 2 mol H 2 O 2 mol H 2 O 2 = 0.50 mol H 2 O 0.50 mol H 2 O 2 O 2 2 mol H 2 O 2 = 0.25 mol O 2 b. 2KClO 3 (s) 2KCl(s) + 3O 2 (g) 0.50 mol KClO 3 2 mol KCl 2 mol KClO 3 = 0.50 mol KCl 0.50 mol KClO 3 3 mol O 2 2 mol KClO 3 = 0.75 mol O 2 c. 2Al(s) + 6HCl(aq) 2AlCl 3 (aq) + 3H 2 (g) 0.50 mol Al 2 mol AlCl 3 2 mol Al = 0.50 mol AlCl mol Al 3 mol H 2 2 mol Al = 0.75 mol H 2

21 94 Chapter 9 d. C 3 H 8 (g) + 5O 2 (g) 3CO 2 (g) + 4H 2 O(l) 0.50 mol C 3 H 8 3 mol CO 2 C 3 H 8 = 1.5 mol CO mol C 3 H 8 4 mol H 2 O C 3 H 8 = 2.0 mol H 2 O 53. a. NH 3 (g) + HCl(g) NH 4 Cl(s) molar mass of NH 3 = 17.0 g 1.00 g NH 3 NH g NH 3 = mol NH mol NH 3 NH 4 Cl NH 3 = mol NH 4 Cl b. CaO(s) + CO 2 (g) CaCO 3 (s) molar mass CaO = 56.1 g CaO 1.00 g CaO = mol CaO 56.1 g CaO mol CaO CaCO 3 CaO = mol CaCO 3 c. 4Na(s) + O 2 (g) 2Na 2 O(s) molar mass Na = g 1.00 g Na Na g Na = mol Na mol Na 2 mol Na 2 O 4 mol Na = mol Na 2 O d. 2P(s) + 3Cl 2 (g) 2PCl 3 (l) molar mass P = g 1.00 g P P = mol P g P mol P 2 mol PCl 3 2 mol P = mol PCl Na 2 O 2 (s) + 2H 2 O(l) 4NaOH(aq) + O 2 (g) molar masses: Na 2 O 2, g; O 2, g 3.25 g Na 2 O 2 Na 2 O g Na 2 O 2 = mol Na 2 O 2

22 Chemical Quantities mol Na 2 O 2 O 2 2 mol Na 2 O 2 = mol O mol O g O 2 O 2 = g O C 2 H 2 (g) + 5O 2 (g) 4CO 2 (g) + 2H 2 O(g) molar masses: C 2 H 2, g; O 2, g 150 g = g g C 2 H 2 C 2 H g C 2 H 2 = mol C 2 H mol C 2 H 2 5 mol O 2 2 mol C 2 H 2 = mol O mol O g O 2 O 2 = g O a. C 2 H 5 OH(l) + 3O 2 (g) 2CO 2 (g) + 3H 2 O(l) molar masses: C 2 H 5 OH, g; O 2, g; CO 2, g 25.0 g C 2 H 5 OH C 2 H 5 OH g C 2 H 5 OH = mol C 2H 5 OH 25.0 g O 2 O g O 2 = mol O 2 Since there is less C 2 H 5 OH present on a mole basis, see if this substance is the limiting reactant mol C 2 H 5 OH 3 mol O 2 C 2 H 5 OH = O 2 From the above calculation, C 2 H 5 OH must not be the limiting reactant (even though there is a smaller number of moles of C 2 H 5 OH present) since more oxygen than is present would be required to react completely with the C 2 H 5 OH present. Oxygen is the limiting reactant mol O 2 2 mol CO 2 3 mol O 2 = mol CO mol CO g CO 2 CO 2 = 22.9 g CO 2 b. N 2 (g) + O 2 (g) 2NO(g) molar masses: N 2, g; O 2, g; NO, g 25.0 g N 2 N g N 2 = mol N 2

23 96 Chapter g O 2 O g O 2 = mol O 2 Since the coefficients of N 2 and O 2 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 O 2 present on a mole basis, O 2 must be the limiting reactant mol O 2 2 mol NO O 2 = mol NO mol NO g NO NO = 46.9 g NO c. 2NaClO 2 (aq) + Cl 2 (g) 2ClO 2 (g) + 2NaCl(aq) molar masses: NaClO 2, g; Cl 2, g; NaCl, g 25.0 g NaClO 2 NaClO g NaClO 2 = mol NaClO g Cl 2 Cl g Cl 2 = mol Cl 2 See if NaClO 2 is the limiting reactant mol NaClO 2 Cl 2 NaClO 2 = mol Cl 2 Since mol of NaClO 2 would require only mol Cl 2 to react completely (and since we have more than this amount of Cl 2 ), then NaClO 2 must indeed be the limiting reactant. 2 mol NaCl mol NaClO 2 = mol NaCl 2 mol NaClO g NaCl mol NaCl = 16.2 g NaCl NaCl d. 3H 2 (g) + N 2 (g) 2NH 3 (g) molar masses: H 2, g; N 2, g; NH 3, g 25.0 g H 2 H g H 2 = mol H g N 2 N g N 2 = mol N 2 See if N 2 is the limiting reactant mol N 2 3 mol H 2 N 2 = mol H 2

24 Chemical Quantities 97 N 2 is clearly the limiting reactant, since there is mol H 2 present (a large excess) mol N 2 2 mol NH 3 N 2 = mol NH mol NH g NH 3 NH 3 = 30.4 g NH N 2 H 4 (l) + O 2 (g) N 2 (g) + 2H 2 O(g) molar masses: N 2 H 4, g; O 2, g; N 2, g; H 2 O, g 20.0 g N 2 H 4 N 2 H g N 2 H 2 = mol N 2 H g O 2 O g O 2 = mol O 2 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 N 2 H 4 as the limiting reactant in the following calculations mol N 2 H 4 N 2 N 2 H 4 = mol N mol N g N 2 N 2 = 17.5 g N mol N 2 H 4 2 mol H 2 O N 2 H 4 = mol H 2 O = 1.25 mol H 2 O mol H 2 O g H 2 O H 2 O = 22.5 g H 2O g theoretical 40 g actual 100 g theoretical = 5.0 g 59. We are concerned with the balanced equation: X HCl 4XCl 3 + 6H 2 We are given the mass of X 4 (248 g). If we can determine the number of moles of X 4, we can determine its molar mass, and therefore its identity. We can do this from the mass of H g H 2 H g H 2 X 4 6 mol H 2 = 1.98 mol X 4 Thus, 248 g X mol X 4 = 125 g/mol (molar mass of X 4 )

25 98 Chapter 9 Element X has an atomic mass of about This is closest to phosphorus (30.97 g/mol). = 31.3 g/mol. 60. We know the following: x + y xy and x + 3z xz 3 5 g 15 g 3 g 18 g 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 P 4 O 10 from the equation: P 4 + 5O 2 P 4 O g P 4 P g P 4 P 4 O 10 P g P 4 O 10 P 4 O 10 = 45.8 g P 4 O g O 2 O g O 2 P 4 O 10 5 mol O g P 4 O 10 P 4 O 10 = 53.2 g P 4 O 10 Thus, the maximum yield of P 4 O 10 is 45.8 g. Determine the yield of H 3 PO 4 from the equation: P 4 O H 2 O 4H 3 PO g P 4 O 10 P 4 O g P 4 O 10 4 mol H 3 PO 4 P 4 O g H 3 PO 4 H 3 PO 4 = 63.2 g H 3 PO g H 2 O H 2 O g H 2 O 4 mol H 3 PO 4 6 mol H 2 O g H 3 PO 4 H 3 PO 4 = 54.4 g H 3 PO 4 Thus, the maximum yield of H 3 PO 4 is 54.4 g. 62. a. Neither is limiting. N 2 2 mol NH 3 N 2 = 2 moles NH 3 3 mol H 2 2 mol NH 3 3 mol H 2 = 2 moles NH 3 b. H 2 is limiting. 3 mol H 2 2 mol NH 3 3 mol H 2 = 2 moles NH 3 c. H 2 is limiting. 3 mol H 2 2 mol NH 3 = 2 moles NH 3 3 mol H 2 d. So, choice d (each would produce the same amount of product) is the correct answer.

26 Chemical Quantities Choice c is correct. Since the molar mass of O 2 is larger than the molar mass of C 2 H 6, an equal mass of each would result in a fewer number of moles of O 2 than C 2 H 6. If this were the case, O 2 must be limiting since it is needed in a greater amount (a 7:2 mol ratio) g CaO CaO g CaO 50.0 g C C g C 5 mol C 2 mol CaO = 4.16 mol C = 2.23 mol C required Leftover C = 4.16 mol 2.23 mol = 1.93 mol C g C = 23.2 g C left over C (CaO is limiting.) g KClO 3 KClO g KClO 3 2 mol KCl 2 mol KClO g KCl KCl = g KCl Percent yield = actual yield theoretical yield 100% = 6.23 g g 100% = 73.3% yield 66. False. Amounts used can vary. However, according to the equation, for every g of N 2 (2 mol) we need g of H 2 (3 mol). So, for example, if we have g of N 2 and 5.5 g of H 2, the H 2 is limiting. If we have g of N 2 and 6.5 g of H 2, the N 2 is limiting. 67. a. 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. b g Na Na g Na 2 mol NaCl 2 mol Na g NaCl NaCl = 50.8 g 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 Cl 2. Since the problem states that the mass of Cl 2 is the same in all containers, each container will have this amount of Cl g Na Na g Na Cl 2 2 mol Na 70.9 g Cl 2 Cl 2 = 61.7 g Cl 2 d. 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 Cl 2 is in the container, we use the amount of Cl 2 (the limiting reactant) to determine the amount of product g Cl 2 Cl g Cl 2 2 mol NaCl Cl g NaCl NaCl = g NaCl = 102 g NaCl e. In part b, sodium is limiting, and g Cl 2 is left over (half of the Cl 2 is left over). In part d, Cl 2 is limiting and 10.0 g Na is left over (50.0 g 40.0 g).

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