CHAPTER 9 CALCULATIONS FROM CHEMICAL EQUATIONS SOLUTIONS TO REVIEW QUESTIONS. = 2 mol PH 3. a g b = 0.4 g PH. = 12 mol H 2 O.

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1 HEINS v4.qxd 1/30/06 1:58 PM Page 95 CHAPTER 9 CALCULATIONS FROM CHEMICAL EQUATIONS SOLUTIONS TO REVIEW QUESTIONS 1. A e ratio is the ratio between the e amounts of two atoms and/or ecules involved in a chemical reaction.. In order to convert grams to es the ar mass of the compound under consideration needs to be determined. 3. The balanced equation is Ca 3 P + 6 H O 3 Ca1OH + PH 3 Correct: 11 Ca 3 P PH 3 1 Ca 3 P = PH 3 Incorrect: 1 g Ca 3 P would produce 0.4 g PH 3 (d) (1 g Ca 3 P )a g b PH 3 a g b = 0.4 g PH 1 Ca 3 P 3 Correct: see equation Correct: see equation (e) Incorrect: Ca 3 P requires 1 H O to produce 4.0 PH 3. 1 Ca 3 P 6 H O 1 Ca 3 P = 1 H O (f) Correct: Ca 3 P will react with 1 H O ( 3 H O are present in excess) and 6 Ca(OH) will be formed. 1 Ca 3 P 3 Ca(OH) 1 Ca 3 P = 6 Ca(OH) (g) (h) Incorrect: (00. g Ca 3 P )a g b 6 H O a 18.0 g b = 119 g H 1 Ca 3 P O The amount of water present (100. g) is less than needed to react with 00. g Ca 3 P. H O is the limiting reactant. Incorrect: water is the limiting reactant g H Oa g b PH 3 6 H O a g b = 6.9 g PH

2 HEINS v4.qxd 1/30/06 1:58 PM Page The balanced equation is CH O + NH 3 HCN + 6 H O Correct HCN Incorrect: 116 O = 10.7 HCN (not 1 HCN) 3 O Correct (d) Incorrect: 11 HCN 6 H O (not 4 H O) HCN = 36 H O (e) Correct O (f) Incorrect: is the limiting reactant HCN 13 O = HCN 3 O (not 3 HCN) 5. The theoretical yield of a chemical reaction is the maximum amount of product that can be produced based on a balanced equation. The actual yield of a reaction is the actual amount of product obtained. 6. You can calculate the percent yield of a chemical reaction by dividing the actual yield by the theoretical yield and multiplying by one hundred

3 HEINS v4.qxd 1/30/06 1:58 PM Page 97 CHAPTER 9 SOLUTIONS TO EXERCISES 1. (d). (d) 3. (d) (e) 15.0 g KNO 3 a g b = 0.47 KNO m NaOHa b = NaOH 1000 m 15.4 * 10 g (NH 4 ) C O 4 a g b = 4.4 (NH 4) C O 4 The conversion is: ml sol g sol g H SO 4 H SO ml solutiona 1.77 g ml b g H SO 4 a 1 g solution g b = 0.37 H SO kg NaHCO 3 a 1000 g ba 1 kg g b = 5.0 NaHCO 3 1 g mg ZnCl a ba 1000 mg g b = 3.85 * 10-3 ZnCl 19.8 * ecules CO 6.0 * 10 3 ecules = 16 CO 150 ml C H 5 OHa g ml 1.55 Fe(OH) 3 a g b = 73 g Fe(OH) kg CaCO 3 a 1000 g b = 1.5 * 10 5 g CaCO kg NH 3 a g b = 179 g NH 3 1 ba g b = 4.3 C H 5 OH 1 17 m HCla 1000 m ba36.46 g b =.6 g HCl ml Br a g ml b = g Br = * 10 3 g Br

4 HEINS v4.qxd 1/30/06 1:58 PM Page (d) ( NiSO 4 )a g b = 1.31 g NiSO 4 ( HC H 3 O )a g b = 3.60 g HC H 3 O (0.75 Bi S 3 )a 514. g b = 373 g Bi S 3 (4.50 * ecules C 6 H 1 O 6 ) 6.0 * 10 3 ecules a 180. g b (e) (75 ml solutiona g ml ba0.00 g K CrO 4 b = 18 g K g solution CrO 4 = 1.35g C 6 H 1 O 6 5. Larger number of ecules: 10.0 g H O or 10.0 g H O Water has a lower ar mass than hydrogen peroxide grams of water has a lower ar mass, contains more es, and therefore more ecules than 10.0 g of H O. 6. Larger number of ecules: 5.0 g HCl or 85 g C 6 H 1 O g HCla g ba6.0 * 103 ecules b = 4.13 * 10 3 ecules HCl g C 6 H 1 O 6 a g ba6.0 * 103 ecules b HCl contains more ecules =.84 * 10 3 ecules C 6 H 1 O 6 7. Mole ratios C 3 H 7 OH + 9 O 6 CO + 8 H O 6 CO C 3 H 7 OH (d) 8 H O C 3 H 7 OH C 3 H 7 OH 9 O (e) 6 CO 8 H O 9 O 6 CO (f) 8 H O 9 O

5 HEINS v4.qxd 1/30/06 1:58 PM Page Mole ratios 3 CaCl + H 3 PO 4 Ca 3 (PO 4 ) + 6 HCl 3 CaCl 1 Ca 3 (PO 4 ) (d) 1 Ca 3 (PO 4 ) H 3 PO 4 6 HCl H 3 PO 4 (e) 6 HCl 1 Ca 3 (PO 4 ) 3 CaCl H 3 PO 4 (f) H 3 PO 4 6 HCl 9. C H 5 OH + 3 O CO + 3 H O CO C H 5 OH 1 C H 5 OH = 15.5 CO 10. The balanced equation is 4 HCl + O Cl + H O HCl Cl 4 HCl =.80 Cl 11. The balanced equation is MnO (s) + 4 HCl(aq) Cl (g) + MnCl (aq) + H O(l) 4 HCl MnO = 4.0 HCl 1 MnO 11.5 H O 1 MnCl H O = 0.65 MnCl s 1. Al 4 C H O 4 Al(OH) CH g Al 4 C 3 a g b 1 H O 1 Al 4 C 3 = 8.33 H O CH 4 4 Al(OH) 3 3 CH 4 = Al(OH)

6 HEINS v4.qxd 1/30/06 1:58 PM Page Grams of NaOH Ca(OH) + Na CO 3 NaOH + CaCO 3 The conversion is: g Ca(OH) Ca(OH) NaOH g NaOH 1500 g Ca(OH) a 1 NaOH ba ba g b = 5 * 10 g NaOH g 1 Ca(OH) 14. Grams of Zn 3 (PO 4 ) 3 Zn + H 3 PO 4 Zn 3 (PO 4 ) + 3 H The conversion is: g Zn Zn Zn 3 (PO 4 ) g Zn 3 (PO 4 ) g Zna g b 1 Zn 3(PO 4 ) a g b = 19.7 g Zn 3 Zn 3 (PO 4 ) 15. The balanced equation is Fe O C Fe + 3 CO The conversion is: kg Fe O 3 k Fe O 3 k Fe kg Fe 115 kg Fe O 3 a 1 k k Fe kg ba ba b = 87.4 kg Fe kg 1 k Fe O 3 k 16. The balanced equation is 3 Fe + 4 H O Fe 3 O H Calculate the grams of both H O and Fe to produce 375 g Fe 3 O g Fe 3 O g a 4 H O a 18.0 g b = 117 g H 1 Fe 3 O 4 O 1375 g Fe 3 O 4 a 1 3 Fe b a g b = 71 g Fe 31.6 g 1 Fe 3 O The balanced equation is C H O 4 CO + 6 H O C H 6 7 O C H 6 = 5.5 O g H Oa g b 4 CO 6 H O a g b = 13.0 g CO g C H 6 a g b 4 CO a g b =.0 * 10 g CO C H 6 (d) 1.75 of H O 4 CO 6 H O a g b = 80.7 g CO

7 HEINS v4.qxd 1/30/06 1:58 PM Page 101 (e) 15.0 C H 6 7 O C H 6 a 3.00 g O b =.80 * 10 3 g O (f) 115 g H O g C H 6 6 H O a g b = 69.5 g C H FeS + 11 O Fe O SO FeS a Fe O 3 4 FeS b = Fe O FeS 11 O 4 FeS = 1.4 O Fe O 3 8 SO Fe O 3 = 6.0 SO (d) FeS 8 SO a g b = 65.6 g SO 4 FeS (e) g SO a g b 11 O 8 SO = O (f) 11 g Fe O 3 a g b 4 FeS a 10.0 g b = 33 g FeS Fe O Hydrogen Oxygen Hydrogen is the limiting reactant

8 HEINS v4.qxd 1/30/06 1:58 PM Page 10 Hydrogen Bromine Bromine is the limiting reactant. 0. Lithium Iodine No limiting reactant. Silver Chlorine Silver is the limiting reactant. 1. Potassium Chlorine Potassium is the limiting reactant

9 HEINS v4.qxd 1/30/06 1:58 PM Page 103 Aluminum Oxygen Oxygen is the limiting reactant.. Nitrogen Oxygen Oxygen is the limiting reactant. Iron Hydrogen Oxygen Water is the limiting reactant. 3. KOH + HNO 3 KNO 3 + H O 16.0 g 1.0 g Choose one of the products and calculate its mass that would be produced from each given reactant. Using KNO 3 as the product: g KOH g a 1 KNO 3 1 KOH a g b = 8.8 g KNO g HNO 3 a g b 1 KNO 3 1 KOH a g b = 19.3 g KNO 3 Since HNO 3 produces less KNO 3, it is the limiting reactant and KOH is in excess

10 HEINS v4.qxd 1/30/06 1:58 PM Page 104 NaOH + H SO 4 Na SO 4 + H O 10.0 g 10.0 g Choose one of the products and calculate its mass that would be produced from each given reactant. Using H O as the product: g NaOHa g b H O NaOH a 18.0 g b = 4.51 g H O g H SO 4 a g b H O a 18.0 g b = 3.67 g H 1 H SO 4 O Since H SO 4 produces less H O, it is the limiting reactant and NaOH is in excess. 4. Bi(NO 3 ) H S Bi S HNO g 6.00 g Choose one of the products and calculate its mass that would be produced from each given reactant. Using Bi S 3 as the product: g Bi(NO 3 ) 3 )a g ba 1 Bi S 3 ba 514. g b = 3.5 g Bi Bi(NO 3 ) 3 S 3 (6.00 g H S)a g b 1 Bi S 3 3 H S a 514. g b = 30. g Bi S 3 Since H S produces less Bi S 3, it is the limiting reactant and Bi(NO 3 ) 3 is in excess. 3 Fe + 4 H O Fe 3 O H 40.0 g 16.0 g Choose one of the products and calculate its mass that would be produced from each given reactant. Using as the product: H (40.0 g Fe)a g b 4 H 3 Fe a.016 g b = 1.93 g H g H Oa g b 4 H 4 H O a.016 g b = 1.79 g H Since H O produces less H, it is the limiting reactant and Fe is in excess. 5. Limiting reactant calculations C 3 H O 3 CO + 4 H O Reaction between 0.0 g C 3 H 8 and 0.0 g O Convert each amount to es of CO

11 HEINS v4.qxd 1/30/06 1:58 PM Page g C 3 H 8 a g b 3 CO 1 C 3 H 8 = 1.36 es CO 10.0 g O a g b 3 CO 5 O = es CO O is the limiting reactant. The yield is es CO. Reaction between 0.0 g C 3 H 8 and 80.0 g O Convert each amount to es of CO 10.0 g C 3 H 8 a g b 3 CO 1 C 3 H 8 = 1.36 es CO g O a g b 3 CO 5 O = 1.50 es CO C 3 H 8 is the limiting reactant. The yield is 1.36 es CO. Reaction between.0 C 3 H 8 and 14.0 O According to the equation, C 3 H 8 will react with 10 O. Therefore, C 3 H 8 is the limiting reactant and 4.0 O will remain unreacted. 1.0 C 3 H 8 3 CO 1 C 3 H 8 = 6.0 CO produced (.0 C 3 H 8 ) 4 H O 1 C 3 H 8 = 8.0 H O produced When the reaction is completed, 6.0 CO, 8.0 H O, and 4.0 O will be in the container. 6. The balanced equation is C 3 H O 6 CO + 6 H O Reaction between 15.0 g C 3 H 6 and 15.0 g O. Convert each amount to es of H O g C 3 H g 6 H O C 3 H 6 = 1.07 H O g O g 6 H O O = H O The O is the limiting reactant. The yield is H O

12 HEINS v4.qxd 1/30/06 1:58 PM Page 106 Reaction between 1.0 g of C 3 H 6 and 5.0 g of O. Convert each amount to es of H O 11.0 g C 3 H g 6 H O C 3 H 6 = H O 15.0 g O g 6 H O 9 O = 0.51 H O O is the limiting reactant. The yield is 0.51 H O. Reaction between 5.0 of C 3 H 6 and 15.0 of O. Convert each to es of CO (5.0 C 3 H 6 ) 6 CO C 3 H 6 = 15 CO (15.0 O ) 6 CO = 10 CO 9 O Since is the limiting reactant. C 3 H 6 will be left unreacted. O X O 8 XO 3 The conversion is: g O O X 8 (10.0 g O a g b 1 X 8 1 O = X g X 8 = X g = 56 g> X 8 g 56 g ar mass X = = Using the periodic table we find that the element with 3.0 g> is sulfur. X + HCl XCl + H The conversion is: g H H X 1.4 g H a g b 1 X 1 H = 1.0 X 78.5 g X = 1.0 X 78.5 g = 65.4 g> 1.0 Using the periodic table we find that the element with atomic mass 65.4 is zinc

13 HEINS v4.qxd 1/30/06 1:58 PM Page Limiting reactant calculation and percentage yield Al + 3 Br AlBr 3 Reaction between 5.0 g Al and 100. g Br Calculate the grams of AlBr 3 from each reactant g Ala g b AlBr 3 a 66.7 g b = 47 g AlBr Al 3 (100. g Br )a g b AlBr 3 a 66.7 g b = 111 g AlBr 3 Br 3 Br is limiting; 111 g AlBr 3 is the theoretical yield of product. actual yield 64. g Percent yield = a b1100 = a b1100 = 57.8% theoretical yield 111 g 30. Percent yield calculation Fe(s) + CuSO 4 (aq) Cu(s) + FeSO 4 (aq) g CuSO 4 a 1 1 Cu b a g b = 159 g Cu (theoretical yield) g 1 CuSO 4 actual yield 151 g % yield = a b1100 = a b1100 = 95.0% yield of Cu theoretical yield 159 g 31. The balanced equation is 3 C + SO CS + CO Calculate the g C needed to produce 950 g CS taking into account that the yield of CS is 86.0%. First calculate the theoretical yield of CS. 950 g CS = 1.1 * 10 3 g CS 1theoretical yield Now calculate the grams of coke needed to produce 1.1 * 10 3 g CS * 10 3 g CS a g b 3 C a 1.01 g b = 5. * 10 g C 1 CS 3. The balanced equation is CaC + H O C H + Ca1OH First calculate the grams of pure CaC in the sample from the amount of C H produced C H 1 CaC a g CaC b = 34.6 g of pure CaC 1 C H CaC in the sample

14 HEINS v4.qxd 1/30/06 1:58 PM Page 108 Now calculate the percent CaC in the impure sample g CaC 44.5 g sample 1100 = 77.8% CaC in the impure sample 33. No. There are not enough screwdrivers, wrenches or pliers. 400 screwdrivers, 3600 wrenches and 100 pliers are needed for 600 tool sets. 34. A subscript is used to indicate the number of atoms in a formula. It cannot be changed without changing the identity of the substance. Coefficients are used only to balance atoms in chemical equations. They may be changed as needed to achieve a balanced equation. 35. Consider the reaction A B and assume that you have 1 gram of A. This does not guarantee that you will produce 1 gram of B because A and B have different ar masses. One gram of A does not contain the same number of ecules as 1 gram of B. However, 1 e of A does have the same number of ecules as one e of B. (Remember, 1 e = 6.0 * 10 3 ecules always.) If you determine the number of es in one gram of A and multiply by to get the number of es of B Á then from that you can determine the grams of B using its ar mass. Equations are written in terms of es not grams KO + H O + 4 CO 4 KHCO O 0.85 g CO a 1 min g b 4 KO = KO 4 CO min KO b110.0 min = 0.19 KO min 37. The conversion is: g CO min CO min O min 0.85 g CO a 1 min g b 3 O a 3.00 g 4 CO 1750 g C 6 H 1 O 6 a g b C H 5 OH a g b = 380 g C 1 C 6 H 1 O 6 H 5 OH ba g O min g O hr 60.0 min b = 8 g O 1.0 hr hr 1750 g C 6 H 1 O 6 a g b CO a g b = 370 g CO 1 C 6 H 1 O 6 Alternate Solution: 750 g C H 6 O g C H 5 OH = 370 g CO by the conservation of mass method g C H 5 OHa 1 ml 0.79 g b = 480 ml C H 5 OH

15 HEINS v4.qxd 1/30/06 1:58 PM Page CH 3 OH + 3 O CO + 4 H O The conversion is: ml CH 3 OH g CH 3 OH CH 3 OH O g O ml CH 3 OHa 0.7 g ml b g 3 O CH 3 OH a 3.00 g b = 65 g O 39. The balanced equation is The conversion is: 7 H O + N H 4 HNO H O 175 kg N H 4 a 1000 g 1 kg b g HNO 3 a 63.0 g b =.9 * 10 5 g HNO 1 N H L H O a 1000 ml 1 L b 1.41 g 1 1 ml 34.0 g a 8 H O ba 18.0 g 7 H O 175 g H O a g b 1 N H 4 a 3.05 g b = 97.6 g N 7 H O H 4 b =.1 * 10 5 g H O (d) Reaction between 750 g of N H and 15 g of H O. Convert each amount to grams of H O g N H 4 a g b 8 H O a 18.0 g b = 3.4 * 10 3 g H 1 N H 4 O 115 g H O a g b 8 H O a 18.0 g b = 75.7 g H 7 H O O 75.7 g H O can be produced. (e) Since H O is the limiting reactant, N H 4 is in excess. 115 g H O a g b 1 N H 4 a 3.05 g b = 16.8 g N 7 H O H 4 reacted 750 g N H 4 given g N H 4 used = 730 g N H 4 remaining

16 HEINS v4.qxd 1/30/06 1:58 PM Page The balanced equation is 16 HCl + KMnO 4 5 Cl + KCl + MnCl + 8 H O Reaction between 5 g KMnO 4 and 85 g HCl. Convert each to es of MnCl. 15 g KMnO 4 a 1 KMnO g KMnO 4 ba MnCl KMnO 4 b = 0.16 MnCl 185 g HCla g ba MnCl 16 HCl b = 0.9 MnCl KMnO 4 is the limiting reactant; 0.16 MnCl produced. 175 g KCla g ba8 H O KCl ba18.0 g b = 73 g H O Theoretical yield is 91 g Cl ; Percent yield: a 75 g b(100) = 8% yield 91 g (d) Reaction between 5 g HCl and 5 g KMnO 4. Convert each amount to grams of CL. (e) 1150 g HCla g ba 5 Cl 16 HCl ba70.90 g b = 91 g Cl 15 g HCla g ba 5 Cl 16 HCl ba70.90 g b = 15 g Cl 15 g KMnO 4 a g ba 5 Cl ba g b = 8 g Cl KMnO 4 HCl is the limiting; KMnO 4 is in excess; 15 g Cl will be produced. Calculate the mass of unreacted KMnO 4 : 15 g HCla g ba KMnO 4 ba g b = 14 g KMnO 16 HCl 4 will react Unreacted KMnO 4 = 5 g -14 g = 11 g KMnO 4 remain unreacted. 41. The balanced equation is 4Ag + H S + O Ag S + H O 11.1 g Aga g ba Ag S 4 Ag ba47.9 g b = 1.3 g Ag S g H Sa g ba Ag S H S ba47.9 g b = 1.0 g Ag S

17 HEINS v4.qxd 1/30/06 1:58 PM Page 111 H S is limiting 1.0 g Ag S forms g g = 0.03 grams more H S needed to completely react Ag. 4. Mass of the beaker g beaker + Ca(OH) g beaker + CaO g H O absorbed g H Oa g b =.5 * 10- H O absorbed Since the reaction is a 1:1 e, the amount of CaO in the beaker is.5 * Convert to grams. 1.5 * g CaO CaOa b = 1.6 g CaO in the beaker g -1.6 g g g O a g ba Ag S ba 47.9 g b = 1. g Ag 1 O S 11.1 g Aga g ba H S 4 Ag ba34.09 g b = 0.17 g H S reacts beaker + CaO CaO mass of the beaker 43. Pb(NO 3 ) (aq) + KI(aq) PbI (s) + KNO 3 (aq) The solid is lead (II) iodide, PbI. Double displacement reaction. Calculate the es of each reactant. [15 g Pb(NO 3 ) ] + a g b = Pb(NO 3) 115 g KIa 1 b = KI g Stoichiometric quantities of reactants are used. Theoretical yield of PbI is Actual yield: Percent yield: a g PbI a g b = PbI b1100 = 3% yield

18 HEINS v4.qxd 1/30/06 1:58 PM Page Composition of a mixture of KNO 3 and KCl. In the mixture only KCl reacts with AgNO 3. KCl(aq) + AgNO 3 (aq) AgCl(s) + KNO 3 (aq) g KCl g AgCla b =.5 g KCl in the mixture g AgCl g mixture -.5 g KCl = 7.75 g KNO 3.5 g KCl a b1100 =.5% KCl g mixture a 7.75 g KNO g mixture b1100 = 77.5% KNO The balanced equation is Zn + HCl ZnCl + H g Zn - 35 g Zn = 145 g Zn reacted with HCl 1145 g Zna g ba1 H 1 Zn b a.016 g b = 4.47 g H produced 1145 g Zna 1 HCl ba g 1 Zn ba36.46 g b = 16 g HCl reacted g Zna 1 HCl ba g 1 Zn ba36.46 g b = 01 g HCl reacts 01 g - 16 g = 39 g more HCl needed to react wih the g Zn 46. Fe(s) + CuSO 4 (aq) Cu(s) FeSO 4 (aq).0 Fe react with.0 CuSO 4 to yield.0 Cu and.0 FeSO CuSO 4 is unreacted. At the completion of the reaction, there will be.0 Cu,.0 FeSO 4, and 1.0 CuSO 4. Determine which reactant is limiting and then calculate the g FeSO 4 produced from that reactant g Fea 1 Cu ba g 1 Fe ba63.55 g b =.8 g Cu g CuSO 4 a 1 1 Cu ba ba g b = 15.9 g Cu g 1 CuSO 4 Since CuSO 4 produces less Cu, it is the limiting reactant. Determine the mass of FeSO 4 produced from 40.0 g CuSO

19 HEINS v4.qxd 1/30/06 1:58 PM Page g CuSO 4 a g b 1 FeSO 4 a g b = 38.1 g FeSO 1 CuSO 4 4 produced Calculate the mass of unreacted Fe g CuSO 4 a g b 1 Fe a g b = 14.0 g Fe will react 1 CuSO 4 Unreacted Fe = 0.0 g g = 6.0 g. Therefore, at the completion of the reaction, 15.9 g Cu, 38.1 g FeSO 4, 6.0 g Fe, and no CuSO 4 remain. 47. Limiting reactant calculation CO(g) + H (g) CH 3 OH(l) Reaction between 40.0 g CO and 10.0 g H : determine the limiting reactant by calculating the amount of CH 3 OH that would be formed from each reactant g COa g b 1 CH 3OH 1 CO a 3.04 g b = 45.8 g CH 3 OH g H a g b 1 CH 3OH a 3.04 g b = 79.5 g CH H 3 OH CO is limiting; H is in excess; 45.8 g CH 3 OH will be produced. Calculate the mass of unreacted H : g COa g b H 1 CO a.016 g b = 5.76 g H react 10.0 g H g H = 4. g H remain unreacted 48. The balanced equation is C 6 H 1 O 6 C H 5 OH + CO First calculate the theoretical yield g C 6 H 1 O 6 a g b C H 5 OH a g b 1 C 6 H 1 O 6 Then take 84.6% of the theoretical yield to obtain the actual yield. actual yield = 1theoretical yield = 3. * 10 g C H 5 OH = 3.8 * 10 g C H 5 OH (theoretical yield) = 13.8 * 10 g C H 5 OH

20 HEINS v4.qxd 1/30/06 1:58 PM Page g C H 5 OH represents 84.6% of the theoretical yield. Calculate the theoretical yield. theoretical yield = 475 g = 561 g C H 5 OH Now calculate the g C 6 H 1 O 6 needed to produce 561 g C H 5 OH g C H 5 OHa g b 1 C 6H 1 O 6 C H 5 OH a 180. g b = 1.10 * 10 3 g C 6 H 1 O The balanced equations are: CaCl (aq) + AgNO 3 (aq) Ca(NO 3 ) (aq) + AgCl(s) MgCl (aq) + AgNO 3 (aq) Mg1NO 3 (aq) + AgCl(s) 1 of each salt will produce the same amount ( ) of AgCl. MgCl has a higher percentage of Cl than CaCl because Mg has a lower atomic mass than Ca. Therefore, on an equal mass basis, MgCl will produce more AgCl than will CaCl. Calculations show that 1.00 g MgCl produces 3.01 g AgCl, and 1.00 g CaCl produces.56 g AgCl. 50. The balanced equation is Li O + H O LiOH The conversion is: g H O H O Li O g Li O kg Li O 500 g H O astronaut day a g b 1 Li O 4.1 kg Li O astronaut day 130 days13 astronauts = 3.7 * 10 kg Li O 51. The balanced equation is H SO 4 + NaCl Na SO 4 + HCl First calculate the g HCl to be produced 10.0 L HCl solutiona Then calculate the g H SO 4 required to produce the HCl * 10 4 g HCla g b 1 H SO 4 a g HCl 1 b = 1.36 * 104 g H SO 4 Finally, calculate the kg H SO 4 (96%) 1 H O a 9.88 g ba 1 kg 1000 g b = 4.1 kg Li O astronaut day 1000 ml ba 1.0 g 1 L 1.00 ml b10.40 = 1.01 * 104 g HCl * 10 4 g H SO g H SO 4 solution 0.96 g H SO 4 a 1 kg 1000 g b = 14 kg concentrated H SO

21 HEINS v4.qxd 1/30/06 1:58 PM Page Percent yield of H SO g Sa 1 b = S to start with 3.07 g S SO - 10% =.806 SO SO.806 SO 3-10% =.55 SO SO 3.55 H SO 4-10% =.73 H SO H SO 4 a g b = 3.0 g H SO 4 formed Sa 1 H SO 4 b = H 1 S SO 4 (theoretical yield).73 H SO H SO = 7.90% yield Alternate Solution: Calculation of yield. There are three chemical steps to the formation of H SO 4. Each step has a 10% loss of yield. Step 1: Step : Step 3: 100% yield - 10% = 90.00% yield 90.00% yield - 10% = 81.00% yield 81.00% yield - 10% = 7.90% yield Now calculate the grams of product. One e of sulfur will yield a maximum of 1 H SO 4. Therefore S will give a maximum of H SO Sa 1 H SO 4 1 S ba g b = 3.0 g H SO 4 yield 53. According to the equations, the es of CO come from both reactions and the es H O come from only the first reaction. So the NaHCO 3 = * H O = * = NaHCO NaHCO 3 a g b = 6.00 g NaHCO 3 in the sample g NaHCO 3 + Na CO g NaHCO 3 = 4.00 g Na CO 3 in the sample

22 HEINS v4.qxd 1/30/06 1:58 PM Page 116 a 6.00 g NaHCO 3 b1100 = 60.0% NaHCO g 3 a 4.00 g Na CO 3 b1100 = 40.0% Na g CO The balanced equation is KClO 3 KCl + 3 O 1.8 g mixture g residue = 3.37 g O lost by heating Because the O lost came only from KClO 3, we can use it to calculate the amount of KClO 3 in the mixture. The conversion is: g O O KClO 3 g KClO g O a g ba KClO 3 ba 1.6 g b = 8.61 g KClO 3 O 3 in the mixture a 8.61 g KClO g sample b1100 = 67.% KClO The balanced equation is Al(OH) 3 (s) + 3 HCl(aq) AlCl 3 (aq) + 3 H O(l) The conversion is: L HCl g HCl HCl Al(OH) 3 : g Al(OH) 3 a.5 L day g HCl ba3.0 ba 1 L g b 1 Al1OH 3 3 HCl Now calculate the number of 400. mg tablets that can be made from 5.3 g Al(OH) g Al(OH) mg a ba 1 tablet b = 13 tablets>day day g 400. mg a g b = 5.3 g Al1OH 3 >day P + 5 O P 4 O 10 P 4 O H O 4 H 3 PO 4 In the first reaction: 10.0 g Pa 1 b = P g g O a g b = O P 3.44 P This is a ratio of = O 5.00 O Therefore, P is the limiting reactant and the P 4 O 10 produced is:

23 HEINS v4.qxd 1/30/06 1:58 PM Page P 1 P 4O 10 4 P = 0.16 P 4O 10 In the second reaction: g H Oa g b = 0.83 H O H O and we have 0.16 P 4 O 10. The ratio of is = P 4 O Therefore, H O is the limiting reactant and the H 3 PO 4 produced is: H O 4 H 3PO 4 6 H O a g b = 54.4 g H 3 PO

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