SLUTINS MANUAL Stoichiometry Section 11.1 Defining Stoichiometry pages 368 37 Practice Problems pages 371 37 1. Interpret the following balanced chemical equations in terms of particles, moles, and mass. Show that the law of conservation of mass is observed. a. N 3 NH 3 1 molecule N 3 molecules molecules NH 3 1 mole N 3 moles moles NH 3 Mass: N : ( mol(14.007 g/mol 8.014 g 3 : (6 mol(1.008 g/mol 6.048 g NH 3 : ( mol(14.007 g/mol (6 mol(1.008 g/mol 34.06 g c. Mg(s Mg(s atoms Mg 1 molecule 0 formula units Mg moles Mg 1 mole moles Mg Mass: Mg: ( mol(4.305 g/mol 48.610 g : ( mol(15.999 g/mol 31.998 g Mg: ( mol(4.305 g/mol ( mol(15.999 g/mol 80.608 g 48.610 g Mg 31.998 g 80.608 g Mg 80.608 g reactants 80.608 g products. Challenge For each of the following, balance the chemical equation; interpret the equation in terms of particles, moles, and mass; and show that the law of conservation of mass is observed. a. Na(s (l NaH(aq 8.014 g N 6.048 g 0 34.06 g NH 3 34.06 g reactants 34.06 g products b. HCl(aq KH(aq KCl(aq (l 1 molecule HCl 1 formula unit KH 1 formula unit KCl 1 molecule 1 mole HCl 1 mole KH 1 mole KCl 1 mole Mass: HCl: (1 mol(1.008 g/mol (1 mol(35.453 g/mol 36.461 g KH: (1 mol(39.098 g/mol (1 mol(15.999 g/mol (1 mol(1.008 g/mol 56.105 g KCl: (1 mol(39.098 g/mol (1 mol(35.453 g/mol 74.551 g : ( mol(1.008 g/mol (1 mol( 15.999 g/mol 18.015 g Na(s (l NaH(aq 1 atoms Na molecules formula units NaH 1 molecule mol Na mol mol NaH 1 mol Mass: Na: ( mol(.990 g/mol 45.980 g : (4 mol(1.008 g/mol ( mol(15.999 g/mol 36.030 g NaH: ( mol(.990 g/mol ( mol(15.999 g/mol ( mol(1.008 g/mol 79.994 g : ( mol(1.008 g/mol.016 g 45.980 g Na 36.030 g 79.994 g NaH.016 g 8.01 g reactants 8.01 g products 36.461 g HCl 56.105 g KH 74.551 g KCl 18.015 g 9.566 g reactants 9.566 g products Solutions Manual Chemistry: Matter and Change Chapter 11 09
SLUTINS MANUAL b. n(s HN (aq n(n (aq N (l 4n(s 10HN (aq 4n(N (aq 1N 5 (l 4 atoms n 10 molecules HN 4 formula units n(n 1 molecule N 5 molecules 4 mol n 10 mol HN 4 mol n(n 1 mol N 5 mol Mass: 4n: (4 mol(65.39 g/mol 61.56 g 10HN : (10 mol(1.008 g/mol (10 mol(14.007 g/mol (30 mol(15.999 g/mol 630.1 g 4n(N : (4 mol(65.39 g/mol (8 mol(14.007 g/mol (4 mol(15.999 g/mol 757.59 g N : ( mol(14.007 g/mol (1 mol(15.999 g/mol 44.013 g 5 : (10 mol(1.008 g/mol (5 mol(15.999 g/mol 90.075 g 61.56 g n 630.1 g HN 757.59 g n(n 44.013 g N 90.075 g 891.68 g reactants 891.68 g products 3. Determine all possible mole ratios for the following balanced chemical equations. a. 4Al(s 3 Al (s 4 mol Al 3 mol mol Al 3 mol mol Al 4 mol Al 3 mol 4 mol Al mol Al 4 mol Al 3 mol mol Al b. 3Fe(s 4 (l Fe 3 4 (s 4 3 mol Fe 3 mol Fe 3 mol Fe 4 mol 4 mol 1 mol Fe 3 4 4 mol 3 mol Fe 4 mol 3 mol Fe 1 mol Fe 3 4 3 mol Fe 4 mol 1 mol Fe 3 4 1 mol Fe 3 4 4 mol 4 mol 4 mol 4 mol 4 mol 4 mol 1 mol Fe 3 4 1 mol Fe 3 4 4 mol c. Hg(s Hg(l mol Hg 1 mol 1 mol mol Hg mol Hg mol Hg mol Hg mol Hg mol Hg mol Hg 1 mol 1 mol 4. Challenge Balance the following equations, and determine the possible mole ratios. a. n(s HCl(aq ncl (aq (l n(s HCl(aq ncl (aq (l 1 mol n 1 mol n 1 mol n mol HCl 1 mol ncl 1 mol mol HCl mol HCl mol HCl 1 mol n 1 mol ncl 1 mol 1 mol ncl 1 mol n 1 mol 1 mol n 1 mol ncl mol HCl 1 mol mol HCl b. butane (C 4 H 10 oxygen carbon dioxide water 1 mol ncl 1 mol 1 mol 1 mol ncl C 4 H 10 13 0 8C 10 (l mol C 4 H 10 mol C 4 H 10 mol C 4 H 10 13 mol 8 mol C 10 mol 13 mol 8 mol C 10 mol mol C 4 H 10 mol C 4 H 10 mol C 4 H 10 10 mol 10 mol 8 mol C 13 mol 8 mol C 13 mol 13 mol 8 mol C 13 mol 10 mol 10 mol 8 mol C 10 Chemistry: Matter and Change Chapter 11 Solutions Manual
SLUTINS MANUAL Section 11.1 Assessment page 37 5. Compare the mass of the reactants and the mass of the products in a chemical reaction, and explain how these masses are related. The coefficients in the balanced equation indicate the molar relationship between each pair of reactants and products. The masses of reactants and products are equal. 10. Model rite the mole ratios for the reaction of hydrogen gas and oxygen gas,. Make a sketch of six hydrogen molecules reacting with the correct number of oxygen molecules. Show the water molecules produced. 6. State how many mole ratios can be written for a chemical reaction involving three substances. n 3, thus (n(n 1 (3( 6 mole ratios 7. Categorize the ways in which a balanced chemical equation can be interpreted. particles (atoms, molecules, formula units, moles, and mass 8. Apply The general form of a chemical reaction is xa yb zab. In the equation, A and B are elements and x, y, and z are coefficients. State the mole ratios for this reaction. xa/yb and xa/zab 6 3 / and / / and / 6 / and / Sketches should show six hydrogen atoms combining with three oxygen atoms to form six water molecules. yb/xa and yb/zab zab/xa and zab/yb 9. Apply Hydrogen peroxide ( decomposes to produce water and oxygen. rite a balanced chemical equation for this reaction, and determine the possible mole ratios. mol / mol, mol /1 mol, mol / mol, mol /1 mol, 1 mol / mol, 1 mol / mol Section 11. Stoichiometric Calculations pages 373 378 Practice Problems pages 375 377 11. Methane and sulfur react to produce carbon disulfide (CS, a liquid often used in the production of cellophane. C S 8 (s CS (l S a. Balance the equation. C S 8 (s CS (l 4 S b. Calculate the moles of CS produced when 1.50 moles of S 8 is used. 1.50 mol S 8 mol CS 3.00 mol CS 1 mol S 8 Solutions Manual Chemistry: Matter and Change Chapter 11 11
SLUTINS MANUAL c. How many moles of S are produced? 1.50 mol S 8 4 mol S 6.00 mol H 1 mol S S 8 1. Challenge Sulfuric acid ( S 4 is formed when sulfur dioxide (S reacts with oxygen and water. a. rite the balanced chemical equation for the reaction. S (l S 4 (aq b. How many moles of S 4 are produced from 1.5 moles of S? 1.5 mol S mol S 4 mol S 1.5 mol S 4 produced c. How many moles of are needed? 1.5 mol S 1 mol mol S 6.5 mol needed 13. Sodium chloride is decomposed into the elements sodium and chlorine by means of electrical energy, as shown below. How much chlorine gas, in grams, is obtained from the process? Electric Na energy NaCl.50 mol Cl? g Step 1: Balance the chemical equation. NaCl(s Na(s Cl Step : Make mole mole conversion. 1 mol Cl.50 mol NaCl mol NaCl 1.5 mol Cl Step 3: Make mole mass conversion. 1.5 mol Cl 70.9 g Cl 88.6 g Cl 1 mol Cl 14. Challenge Titanium is a transition metal used in many alloys because it is extremely strong and lightweight. Titanium tetrachloride (TiCl 4 is extracted from titanium oxide (Ti using chlorine and coke (carbon. Ti (s C(s Cl TiCl 4 (s C a. hat mass of Cl gas is needed to react with 1.5 mol of Ti? Step 1: Make mole mole conversion. 1.5 mol Ti mol Cl.50 mol Cl 1 mol Ti Step : Make mole mass conversion..50 mol Cl 70.90 g Cl 177 g Cl 1 mol Cl b. hat mass of C is needed to react with 1.5 mol of Ti? Step 1: Make mole mole conversion. 1 mol C 1.5 mol Ti 1.5 mol C 1 mol Ti Step : Make mole mass conversion. 1.5 mol C 1.011 g C 1 mol C 15.0 g C c. hat is the mass of all of the products formed by reaction with 1.5 mol of Ti? Calculate the mass to Ti used. 1.5 mol Ti 79.865 g Ti 99.8 g Ti 1 mol Ti Calculate the total mass of the reactants. 99.8 g 177 g 15.0 g 9 g Because mass is conserved, the mass of the products must equal the mass of reactants. mass of products 9 g 1 Chemistry: Matter and Change Chapter 11 Solutions Manual
SLUTINS MANUAL 15. ne of the reactions used to inflate automobile air bags involves sodium azide (NaN 3 : NaN 3 (s Na(s 3N. Determine the mass of N produced from the decomposition of NaN 3 shown below. N gas 100.0 g NaN 3? g N NaN 3 (s Na(s 3N Step 4: Make mole mass conversion. 0.0390 mol S 4 98.09 g S 4 1 mol S 4 3.83 g S 4 Section 11. Assessment page 378 17. Explain why a balanced chemical equation is needed to solve a stoichiometric problem. The coefficients in the balanced equation indicate the molar relationship between each pair of reactants and products. Step 1: Make mass mole conversion. 100.0 g NaN 3 1 mol NaN 3 1.538 mol NaN 65.0 g NaN 3 3 Step : Make mole mole conversion. 3 mol N 1.538 mol NaN 3.307 mol N mol NaN 3 Step 3: Make mole mass conversion..307 mol N 8.0 g N 64.64 g N 1 mol N 16. Challenge In the formation of acid rain, sulfur dioxide (S reacts with oxygen and water in the air to form sulfuric acid ( S 4. rite the balanced chemical equation for the reaction. If.50 g of S reacts with excess oxygen and water, how much S 4, in grams, is produced? Step 1: Balance the chemical equation. S (l 0 S 4 (aq Step : Make mass mole conversion..50 g S 1 mol S 0.0390 mol S 64.07 g S Step 3: Make mole mole conversion. 0.0390 mol S mol S 4 mol S 18. List the four steps used in solving stoichiometric problems. 1. Balance the equation.. Convert the mass of the known substance to moles of known substance. 3. Use the mole ratio to convert from moles of the known to moles of the unknown. 4. Convert moles of unknown to mass of the unknown. 19. Describe how a mole ratio is correctly expressed when it is used to solve a stoichiometric problem. moles of unknown/moles of known 0. Apply How can you determine the mass of liquid bromine (Br needed to react completely with a given mass of magnesium? rite a balanced equation. Convert the given mass of magnesium to moles. Use the mole ratio from the balanced equation to convert moles of magnesium to moles of bromine. Convert from moles of bromine to mass of bromine. 0.0390 mol S 4 Solutions Manual Chemistry: Matter and Change Chapter 11 13
SLUTINS MANUAL 1. Calculate Hydrogen reacts with excess nitrogen as follows: N 3 NH 3 If.70 g of reacts, how many grams of NH 3 is formed? (.70 g ( 1 mol.016 g 1.34 mol 1.34 mol ( mol NH 3 3 mol 0.893 mol NH 3 (0.893 mol NH 3 ( 17.031 g NH 3 1 mol NH 3 15. g NH 3. Design a concept map for the following reaction. Make mole ratio comparison. 0.661 mol Fe compared to 1 mol Fe 4.350 mol Na 6 mol Na 0.1439 compared to 0.1667 The actual ratio is less than the needed ratio, so iron(iii oxide is the limiting reactant. b. reactant in excess Sodium is the excess reactant. c. mass of solid iron produced Make mole mole conversion. mol Fe 0.661 mol Fe 1 mol Fe 1.5 mol Fe CaC (s HCl(aq CaCl (aq (l C The concept map should explain how to determine the mass of CaCl produced from a given mass of HCl. Concept maps will vary, but all should show the use of these conversion factors: the inverse of molar mass, the mole ratio, the molar mass. Section 11.3 Limiting Reactants pages 379 384 Practice Problems page 383 3. The reaction between solid sodium and iron(iii oxide is one in a series of reactions that inflates an automobile airbag: 6Na(s Fe (s 3Na (s Fe(s. If 100.0 g of Na and 100.0 g of Fe are used in this reaction, determine the following. a. limiting reactant Make mass mole conversion. 1 mol Na 100.0 g Na 4.350 mol Na.99 g Na 100.0 g Fe 1 mol Fe 159.7 g Fe 0.661 mol Fe Make mole mass conversion. 55.85 g Fe 1.5 mol Fe 69.9 g Fe 1 mol Fe d. mass of excess reactant that remains after the reaction is complete Make mole mole conversion. 6 mol Na 0.661 mol Fe 1 mol Fe 3.757 mol Na needed Make mole mass conversion..9 g Na 3.757 mol Na 1 mol Na 86.37 g Na needed 100.0 g Na given 86.37 g Na needed 13.6 g Na in excess 4. Challenge Photosynthesis reactions in green plants use carbon dioxide and water to produce glucose (C 6 H 1 6 and oxygen. A plant has 88.0 g of carbon dioxide and 64.0 g of water available for photosynthesis. a. rite the balanced chemical equation for the reaction. 6C 6 (l 0 C 6 H 1 6 (aq 6 14 Chemistry: Matter and Change Chapter 11 Solutions Manual
SLUTINS MANUAL b. Determine the limiting reactant. Make mass mole conversion. 88.0 g C 1 mol C.00 mol C 44.01 g C 64.0 g 1 mol 18.0 g 3.55 mol Make mole ratio comparison..00 mol C 3.55 mol compared to 6 mol C 6 mol ; 0.563 compared to 1.00 The actual ratio is less than the needed ratio, so carbon dioxide is the limiting reactant. c. Determine the excess reactant. ater is the excess reactant. d. Determine the mass in excess. Make mole mole conversion..00 mol C 6 mol.00 mol H 6 mol C Make mole mass conversion..00 mol 18.0 g 1 mol 36.0 g needed 64.0 g given 36.0 g needed 8.0 g in excess e. Determine the mass of glucose produced. Make mole mole conversion..00 mol C 1 mol C 6 H 1 6 6 mol C 0.333 mol C 6 H 1 6 Make mole 0 mass conversion. 0.333 mol C 6 H 1 6 180.4 g C 6 H 1 6 1 mol C 6 H 1 6 60.0 g C 6 H 1 6 Section 11.3 Assessment page 384 5. Describe the reason why a reaction between two elements comes to an end. one of the reactants is used up 6. Identify the limiting and the excess reactant in each reaction. a. ood burns in a campfire. The wood limits. xygen is in excess. The fire will burn only while wood is present. b. Airborne sulfur reacts with the silver plating on a teapot to produce tarnish (silver sulfide. Silver is the limiting reactant. Sulfur is in excess. hen a layer of tarnish covers the silver surface, it prevents the sulfur in the air from reacting. c. Baking powder in batter decomposes to produce carbon dioxide. A decomposition reaction usually has only one reactant. The reaction is limited by the amount of baking powder present. 7. Analyze Tetraphosphorous trisulphide (P 4 S 3 is used in the match heads of some matches. It is produced in the reaction 8P 4 3S 8 8P 4 S 3. Determine which of the following statements are incorrect, and rewrite the incorrect statements to make them correct. a. 4 mol P 4 reacts with 1.5 mol S 8 to form 4 mol P 4 S 3. correct b. Sulfur is the limiting reactant when 4 mol P 4 and 4 mol S 8 react. Phosphorus is the limiting reactant. c. 6 mol P 4 reacts with 6 mol S 8, forming 130 g P 4 S 3. correct Solutions Manual Chemistry: Matter and Change Chapter 11 15
SLUTINS MANUAL Section 11.4 Percent Yield pages 385 388 Practice Problems page 387 8. Aluminum hydroxide (Al(H 3 is often present in antacids to neutralize stomach acid (HCl. The reaction occurs as follows: Al(H 3 (s 3HCl(aq AlCl 3 (aq 3 (l. If 14.0 g of Al(H 3 is present in an antacid tablet, determine the theoretical yield of AlCl 3 produced when the tablet reacts with HCl. Make mass mole conversion. 14.0 g Al(H 3 1 mol Al(H 3 78.0 g Al(H 3 0.179 mol Al(H 3 Make mole mole conversion. 1 mol AlCl 3 0.179 mol Al(H 3 1 mol Al(H 3 0.179 mol AlCl 3 Make mole mass conversion. 0.179 mol AlCl 3 133.3 g AlCl 3 3.9 g AlCl 1 mol AlCl 3 3 3.9 g of AlCl 3 is the theoretical yield. 9. inc reacts with iodine in a synthesis reaction: n I 0 ni. a. Determine the theoretical yield if 1.91 mol of zinc is used. rite the balanced chemical equation. n(s I (s ni (s Make mole mole conversion. 1.91 mol n 1 mol ni 1 mol n 1.91 mol ni Make mole mass conversion. 1.91 mol ni 319. g ni 610.3 g ni 1 mol ni 610.3 g of ni is the theoretical yield. b. Determine the percent yield if 515.6 g of product is recovered. % yield 515.6 g ni 100 610.3 g ni 84.48% yield of ni 30. Challenge hen copper wire is placed into a silver nitrate solution (AgN, silver crystals and copper(ii nitrate (Cu(N solution form. a. rite the balanced chemical equation for the reaction. Cu(s AgN (aq Ag(s Cu(N (aq b. If a 0.0g sample of copper is used, determine the theoretical yield of silver. Make mass mole conversion. 1 mol Cu 0.0 g Cu 0.315 mol Cu 63.55 g Cu Make mole mole conversion. mol Ag 0.315 mol Cu 0.630 mol Ag 1 mol Cu Make mole mass conversion. 107.9 g Ag 0.630 mol Ag 68.0 g Ag 1 mol Ag 68.0 g of Ag is the theoretical yield. c. If 60.0 g of silver is recovered from the reaction, determine the percent yield of the reaction. 60.0 g Ag % yield 68.0 g Ag 100 88.% yield of Ag Section 11.4 Assessment page 388 31. Identify which type of yield theoretical yield, actual yield, or percent yield is a measure of the efficiency of a chemical reaction. percent yield 16 Chemistry: Matter and Change Chapter 11 Solutions Manual
SLUTINS MANUAL 3. List several reasons why the actual yield from a chemical reaction is not usually equal to the theoretical yield. Not all reactions go to completion. Some of the reactants or products stick to the surface of the container and are not massed or transferred. ther unexpected products form from competing reactions. 33. Explain how percent yield is calculated. divide the actual yield by the theoretical yield and multiply the quotient by 100 34. Apply In an experiment, you combine 83.77 g of iron with an excess of sulfur and then heat the mixture to obtain iron(iii sulfide. Fe(s 3S(s Fe S 3 (s Chapter 11 Assessment pages 39 397 Section 11.1 Mastering Concepts 36. hy must a chemical equation be balanced before you can determine mole ratios? Mole ratios are determined by the coefficients in a balanced equation. If the equation is not balanced, the relationship between reactants and products cannot be determined. 37. hat relationships can be determined from a balanced chemical equation? The relationships among particles, moles, and mass for all reactants and products. hat is the theoretical yield, in grams, of iron(iii sulfide? (83.77 g Fe ( 1 mol Fe 1.500 mol Fe 55.845 g Fe 1.500 mol Fe ( 1 mol Fe S 3 155.9 g Fe S 3 mol Fe ( 07.885 g Fe S 3 1 mol Fe S 3 35. Calculate the percent yield of the reaction of magnesium with excess oxygen: Mg(s Mg(s Reaction Data Mass of empty crucible Mass of crucible and Mg Mass of crucible and Mg (after heating 35.67 g 38.06 g 39.15 g mass of Mg 38.06 g 35.67 g.39 g mass of Mg 39.15 g 33.67 g 3.48 g 1 mol Mg Theoretical yield:.39 g Mg 4.31 g Mg 0.0983 mol Mg mol Mg 0.983 mol Mg 40.31 g Mg mol Mg 1 mol Mg 3.96 g Mg 3.48 g Mg % yield 3.96 g Mg 100 87.9% yield of Mg 38. Explain why mole ratios are central to stoichiometric calculations. Mole ratios allow for the conversion from moles of one substance in a balanced chemical equation to moles of another substance in the same equation. 39. hat is the mole ratio that can convert from moles of A to moles of B? moles B moles A 40. hy are coefficients used in mole ratios instead of subscripts? The coefficients in the balanced chemical equation show the numbers of representative particles involved in a reaction. Subscripts give the numbers of different kinds of atoms within a molecule or formula unit. 41. Explain how the conservation of mass allows you to interpret a balanced chemical equation in terms of mass. The mass of the reactants will always equal the mass of the products. Solutions Manual Chemistry: Matter and Change Chapter 11 17
SLUTINS MANUAL 4. hen heated by a flame, ammonium dichromate decomposes, producing nitrogen gas, solid chromium(iii oxide, and water vapor. (N Cr 7 N Cr 4 rite the mole ratios for this reaction that relate ammonium chromate to the products. 1 mol (N Cr 7 1 mol (N Cr 7 1 mol N 1 mol Cr 1 mol (N Cr 7 4 mol 1 mol N 1 mol Cr 1 mol (N Cr 7 1 mol (N Cr 7 4 mol 1 mol (N Cr 7 Mastering Problems 44. Interpret the following equation in terms of particles, moles, and mass. 4Al(s 3 Al (s Particles: 4 atoms Al 3 molecules formula units Al Moles: 4 mol Al 3 mol mol Al ; Mass: 4Al 4 mol Al(6.98 g/mol 107.93 g 3 6 mol (15.999 g/mol 95.99 g Al 4 mol Al(6.98 g/mol 6 mol (15.999 g/mol 03.9 g 107.93 g Al 95.99 g 03.9 g Al 45. Smelting hen tin(iv oxide is heated with carbon in a process called smelting, the element tin can be extracted. Sn (s C(s Sn(l C 43. Figure 11.10 depicts an equation with squares representing Element M and circles representing Element N. rite a balanced equation to represent the picture shown, using smallest whole-number ratios. rite mole ratios for this equation. M N M 4 N mol M N, mol M N, 1 mol M 4 1 mol N 1 mol M 4, 1 mol M 4 1 mol N mol M N, 1 mol N mol M N, 1 mol N 1 mol M 4 Interpret the chemical equation in terms of particles, moles, and mass. Particles: 1 formula unit Sn atoms C 1 atom Sn molecules C Moles: 1 mol Sn mol C 1 mol Sn mol C Mass: Sn : 1 mol(118.710 g/mol mol (15.999 g/mol 150.71 g C: mol(1.011 g/mol 4.0 g Sn: 1 mol(118.710 g/mol 118.71 g C: mol(1.011 g/mol mol (15.999 g/mol 56.0 g 150.71 g Sn 4.0 g C 118.71 g Sn 56.0 g C 18 Chemistry: Matter and Change Chapter 11 Solutions Manual
SLUTINS MANUAL 46. hen solid copper is added to nitric acid, copper(ii nitrate, nitrogen dioxide, and water are produced. rite the balanced chemical equation for the reaction. List six mole ratios for the reaction. Cu(s 4HN (aq Cu(N (aq N (l; answers should include any six of the following ratios: 1 mol Cu/4 mol HN, 1 mol Cu/1 mol Cu(N, 1 mol Cu/ mol N, 1 mol Cu/ mol, 4 mol HN /1 mol Cu, 4 mol HN /1 mol Cu(N, 4 mol HN / mol N, 4 mol HN / mol, 1 mol Cu(N /1 mol Cu, 1 mol Cu(N /4 mol HN, 1 mol Cu(N / mol N, 1 mol Cu(N / mol, mol N /1 mol Cu, mol N /4 mol HN, mol N /1 mol Cu(N, mol N / mol, mol /1 mol Cu, mol /4 mol HN, mol /1 mol Cu(N, mol / mol N 47. hen hydrochloric acid solution reacts with lead(ii nitrate solution, lead(ii chloride precipitates and a solution of nitric acid is produced. a. rite the balanced chemical equation for the reaction. HCl(aq Pb(N (aq PbCl (s HN (aq b. Interpret the equation in terms of molecules and formula units, moles, and mass. Particles: molecules HCl 1 formula unit Pb(N 1 formula unit PbCl molecules HN Moles: mol HCl 1 mol Pb(N 1 mol PbCl mol HN ; Mass: HCl: mol(1.008 g/mol mol (35.453 g/mol 7.9 g Pb(N : 1 mol(07. g/mol mol (14.007 g/mol 6 mol(15.999 g/mol 331. g PbCl : 1 mol(07. g/mol mol (35.453 g/mol 78.1 g 48. hen aluminum is mixed with iron(iii oxide, iron metal and aluminum oxide are produced along with a large quantity of heat. hat mole ratio would you use to determine moles of Fe if moles of Fe are known? Fe (s Al(s Fe(s Al (s heat mol Fe 1 mol Fe 49. Solid silicon dioxide, often called silica, reacts with hydrofluoric acid (HF solution to produce the gas silicon tetrafluoride and water. a. rite the balanced chemical equation for the reaction. Si (s 4HF(aq SiF 4 (l b. List three mole ratios, and explain how you would use them in stoichiometric calculations. Students may write any of 1 ratios. Examples might be the following. 4 mol HF/1 mol Si ; used to find the amount of HF that will react with a known amount of Si ; 1 mol SiF 4 /1 mol Si ; used to find the amount of SiF 4 that can be formed from a known amount of Si. mol /1 mol SiF 4 ; used to find the amount of that will be produced with the SiF 4 50. Chrome The most important commercial ore of chromium is chromite (FeCr 4. ne of the steps in the process used to extract chromium from the ore is the reaction of chromite with coke (carbon to produce ferrochrome (FeCr. C(s FeCr 4 (s FeCr (s C hat mole ratio would you use to convert from moles of chromite to moles of ferrochrome? 1 mol FeCr 1 mol FeCr 4 HN : mol(1.008 g/mol mol (14.007 g/mol 6 mol(15.999 g/mol 16.0 g 7.9 g HCl 331. g Pb(N 78.1 g PbCl 16.0 g HN Solutions Manual Chemistry: Matter and Change Chapter 11 19
SLUTINS MANUAL 51. Air Pollution The pollutant S is removed from the air in a reaction that also involves calcium carbonate and oxygen. The products of this reaction are calcium sulfate and carbon dioxide. Determine the mole ratio you would use to convert moles of S to moles of CaS 4. S CaC CaS 4 C mol CaS 4 mol S 5. Two substances, and X, react to form the products Y and. Table 11. shows the moles of the reactants and products involved when the reaction was carried out. Use the data to determine the coefficients that will balance the equation X Y. Reaction Data Moles of Reactants Moles of Products X Y 0.90 0.30 0.60 1.0 Divide each molar quantity by 0.30 mol, the least common denominator: : 0.90/0.30 3 X: 0.30/0.30 1 Y: 0.60/0.30 : 1.0/0.30 4 3 X 0 Y 4 53. Antacids Magnesium hydroxide is an ingredient in some antacids. Antacids react with excess hydrochloric acid in the stomach to relieve indigestion. Mg(H HCl MgCl a. Balance the reaction of Mg(H with HCl. 1Mg(H HCl 1MgCl b. rite the mole ratio that would be used to determine the number of moles of MgCl produced when HCl reacts with Mg(H. 1 mol MgCl or 1 mol MgCl 1 mol Mg(H mol HCl Section 11. Mastering Concepts 54. hat is the first step in all stoichiometric calculations? rite a balanced chemical equation for the reaction. 55. hat information does a balanced equation provide? The balanced equation provides the relationship between reactants and products, and the coefficients in the equation are used to write mole ratios relating reactants and products. 56. n what law is stoichiometry based, and how do the calculations support this law? Stoichiometry is based on the law of conservation of mass. The calculations are used to determine the mass of reactants and products. nce found, the sum of reactants will equal the sum of products, verifying the law of conservation of mass. 57. How is molar mass used in some stoichiometric calculations? Molar mass is a conversion factor for converting moles of a given substance to mass or mass of a given substance to moles. 58. hat information must you have in order to calculate the mass of product formed in a chemical reaction? You must have the balanced chemical equation and know the quantity of one substance in the reaction other than the product you are to determine. 0 Chemistry: Matter and Change Chapter 11 Solutions Manual
SLUTINS MANUAL + 61. elding If 5.50 mol of calcium carbide (CaC reacts with an excess of water, how many moles of acetylene (C, a gas used in welding, will be produced? 59. Each box in Figure 11.11 represents the contents of a flask. ne flask contains hydrogen sulfide, and the other contains oxygen. hen the contents of the flasks are mixed, a reaction occurs and water vapor and sulfur are produced. In the figure, the red circles represent oxygen, the yellow circles represent sulfur, and blue circles represent hydrogen. a. rite the balanced chemical equation for the reaction. S S(s b. Using the same color code, sketch a representation of the flask after the reaction occurs. Student sketches should show the formation of six water molecules ( and six sulfur atoms (S. Mastering Problems 60. Ethanol (C H, also known as grain alcohol, can be made from the fermentation of sugar (C 6 H 1 6. The unbalanced chemical equation for the reaction is shown below. C 6 H 1 6 C H C Balance the chemical equation and determine the mass of C H produced from 750 g of C 6 H 1 6. C 6 H 1 6 C H C 750 g C 6 H 1 6 (180.16 g/mol 4. mol C 6 H 1 6 4. mol C 6 H 1 6 ( mol C H 1 mol C 6 H 1 6 8.4 mol C H 8.4 mol C H 46.07 g ( 1 mol C H 390 g C H CaC (s (l Ca(H (aq C The mole ratio of CaC :C is 1:1; thus, 5.50 mol C will be produced from 5.50 mol CaC. 6. Antacid Fizz hen an antacid tablet dissolves in water, the fizz is due to a reaction between sodium hydrogen carbonate (NaHC, also called sodium bicarbonate, and citric acid (H 3 C 6 7. 3NaHC (aq H 3 C 6 7 (aq 3C 3 (l Na 3 C 6 7 (aq How many moles of Na 3 C 6 7 can be produced if one tablet containing 0.0119 mol of NaHC is dissolved? 0.0119 mol NaHC (1 mol Na 3 C 6 7 /3 mol NaHC 0.00397 mol 63. Esterification The process in which an organic acid and an alcohol that forms as ester and water is known as esterification. Ethyl butanoate (C 3 H 7 CC, an ester, is formed when the alcohol ethanol (C H and butanoic acid (C 3 H 7 CH are heated in the presence of sulfuric acid. C H(l C 3 H 7 CH(l C 3 H 7 CC (l (l Determine the mass of ethyl butanoate produced if 4.50 mol of ethanol is used. 4.50 mol C H 1 mol C 3 H 7 CC 1 mol C H 116.18 g C 3 H 7 CC 1 mol C 3 H 7 CC 53 g C 3 H 7 CC Solutions Manual Chemistry: Matter and Change Chapter 11 1
SLUTINS MANUAL 64. Greenhouse Gas Carbon dioxide is a greenhouse gas that is linked to global warming. It is released into the atmosphere through the combustion of octane (C 8 H 18 in gasoline. rite the balanced chemical equation for the combustion of octane and calculate the mass of octane needed to release 5.00 mol of C. C 8 H 18 (l 5 0 16C 18 (l 5.00 mol C mol C 8 H 18 0.65 mol C 16 mol C 8 H 18 0.65 mol C 8 H 18 114.8 g C 8 H 18 71.4 g C 1 mol C 8 H 8 H 18 18 65. A solution of potassium chromate reacts with a solution of lead(ii nitrate to produce a yellow precipitate of lead(ii chromate and a solution of potassium nitrate. a. rite the balanced chemical equation. K Cr 4 (aq Pb(N (aq 0 PbCr 4 (s KN (aq b. Starting with 0.50 mol of potassium chromate, determine the mass of lead chromate formed. 0.50 mol K Cr 4 1 mol PbCr 4 1 mol K Cr 4 33. g PbCr 4 80.8 g PbCr 1 mol PbCr 4 4 66. Rocket Fuel The exothermic reaction between liquid hydrazine (N and liquid hydrogen peroxide ( is used to fuel rockets. The products of this reaction are nitrogen gas and water. a. rite the balanced chemical equation. N (l (l 0 N b. How much hydrazine, in grams, is needed to produce 10.0 mol of nitrogen gas? 10.0 mol N 1 mol N 30.03 g N 1 mol N 1 mol N 67. Chloroform (CHCl 3, an important solvent, is produced by a reaction between methane and chlorine. C 3Cl 0 CHCl 3 3HCl How much C in grams is needed to produce 50.0 grams of CHCl 3? 1 mol CHCl 3 50.0 g CHCl 3 1 mol C 119.37 g CHCl 3 1 mol CHCl 3 16.04 g C 6.7 g CH 1 mol C 4 68. xygen Production The Russian Space Agency uses potassium superoxide (K for the chemical oxygen generators in their space suits. 4K 4C 0 4 KHC 3 Complete Table 11.3. xygen Generation Reaction Data Mass Mass Mass Mass Mass K C KHC 1100 g 140 g 7.0 10 g 1600 g 380 g 380 g (1 mol /3.00 g (4 mol K /3 mol (71.1 g K /1 mol K 1100 g 380 g (1 mol /3.00 g ( mol /3 mol (18.0 g /1 mol 140 g 380 g (1 mol /3.00 g (4 mol C /3 mol (44.01 g C /1 mol C 7.0 10 g 380 g (1 mol /3.00 g (4 mol KHC /3 mol (100.1 g KHC /1 mol KHC 1600 g 3.00 10 g N Chemistry: Matter and Change Chapter 11 Solutions Manual
SLUTINS MANUAL 69. Gasohol is a mixture of ethanol and gasoline. Balance the equation, and determine the mass of C produced from the combustion of 100.0 g of ethanol. C H(l 0 C C H(l 3 0 C 3 (l 100.0 g C H 1 mol C H 46.08 g C H.170 mol C H mol C.170 mol C H 1 mol C H 4.340 mol C 4.340 mol C 44.01 g C 1 mol C 191.0 g C produced 70. Car Battery Car batteries use lead, lead(iv oxide, and a sulfuric acid solution to produce an electric current. The products of the reaction are lead(ii sulfate in solution and water. a. rite the balanced chemical equation for this reaction. Pb(s Pb (s S 4 (aq 0 PbS 4 (aq (l b. Determine the mass of lead(ii sulfate produced when 5.0 g of lead reacts with an excess of lead(iv oxide and sulfuric acid. 1 mol Pb 5 g Pb 07. g Pb mol PbS 4 1 mol Pb 303.3 g PbS 4 73. g PbS 1 mol PbS 4 4 71. To extract gold from its ore, the ore is treated with sodium cyanide solution in the presence of oxygen and water. 4Au(s 8NaCN(aq (l 0 4NaAu(CN (aq 4NaH(aq a. Determine the mass of gold that can be extracted if 5.0 g of sodium cyanide is used. 1 mol NaCN 5.0 g NaCN 49.01 g NaCN 4 mol Au 8 mol NaCN 196.97 g Au 50. g Au 1 mol Au b. If the mass of the ore from which the gold was extracted is 150.0 g, what percentage of the ore is gold? 50. g Au 100 33.5% gold in ore 150.0 g ore 7. Film Photographic film contains silver bromide in gelatin. nce exposed, some of the silver bromide decomposes, producing fine grains of silver. The unexposed silver bromide is removed by treating the film with sodium thiosulfate. Soluble sodium silver thiosulfate (Na 3 Ag(S is produced. AgBr(s Na S (aq 0 Na 3 Ag(S (aq NaBr(aq Determine the mass of Na 3 Ag(S produced if 0.75 g of AgBr is removed. 1 mol AgBr 0.75 g AgBr 187.77 g AgBr 1.46 10 3 mol AgBr 1.46 10 3 mol AgBr 1 mol Na 3 Ag(S 1 mol AgBr 1.46 10 3 mol Na 3 Ag(S 1.46 10 3 mol Na 3 Ag(S 401.1 g Na 3 Ag(S 0.587 g Na 1 mol Na 3 Ag(S 3 Ag(S Section 11.3 Mastering Concepts 73. How is a mole ratio used to find the limiting reactant? The actual mole ratio of reactants from the chemical equation is compared to the mole ratio determined from the given quantities. 74. Explain why the statement the limiting reactant is the reactant with the lowest mass is incorrect. The limiting reactant is the reactant that produces the lowest number of moles of product. Mass does not determine the limiting reactant, but the number of moles. Solutions Manual Chemistry: Matter and Change Chapter 11 3
SLUTINS MANUAL 77. Nickel-Iron Battery In 1901, Thomas Edison invented the nickel-iron battery. The following reaction takes place in the battery. 75. Figure 11.1 uses squares to represent Element M and circles to represent Element N. a. rite the balanced equation for the reaction. 3M N 0 M 3 N b. If each square represents 1 mol of M and each circle represents 1 mol of N, how many moles of M and N were present at the start of the reaction? 6 moles of element M (in the form of 3 moles of M and 6 moles of element N (likewise, 3 moles of N c. How many moles of product form? How many moles of M and N are unreacted? moles of M 3 N form with moles of N unreacted (4 total moles of element N d. Identify the limiting reactant and excess reactant. M is the limiting reactant and N is the excess reactant. Mastering Problems 76. The reaction between ethyne (C and hydrogen ( is illustrated in Figure 11.13. The product is ethane (C H 6. hich is the limiting reactant? hich is the excess reactant? Explain. + + Ethyne Hydrogen Ethane Ethyne Hydrogen is limiting; ethyne is the excess reactant. ne mol of ethyne is left over. Fe(s Ni(H(s (l 0 Fe(H (s Ni(H (aq How many mol of Fe(H are produced when 5.00 mol of Fe and 8.00 mol of Ni(H react? According to the balanced equation, two moles of Ni(H react with each mole of Fe. So, 4 mol Fe react with 8.00 mol Ni(H, leaving 1.00 mol Fe in excess. For each mole of Fe that reacts, one mole of Fe(H (s is produced. Because 4.0 mol Fe reacts, 4.0 mol Fe(H is produced. 78. ne of the few xenon compounds that form is cesium xenon heptafluoride (CsXeF 7. How many moles of CsXeF 7 can be produced from the reaction of 1.5 mol of cesium fluoride with 10.0 mol of xenon hexafluoride? CsF(s XeF 6 (s 0 CsXeF 7 (s. 10.0 mol XeF 6 1 mol CsXeF 7 10.0 mol CsXeF 1 mol XeF 7 6 79. Iron Production Iron is obtained commercially by the reaction of hematite (Fe with carbon monoxide. How many grams of iron are produced if 5.0 mol of hematite react with 30.0 mol of carbon monoxide? Fe (s 3C 0 Fe(s 3C According to the balanced equation, 1 mole of hematite reacts with 3 moles of carbon monoxide. Since 5.0 mol of hematite would require 75.0 mol C but only 30.0 mol are available, C is the limiting reactant. mol Fe 30.0 mol C 55.85 g Fe 3 mol C 1 mol Fe 110 g Fe 4 Chemistry: Matter and Change Chapter 11 Solutions Manual
SLUTINS MANUAL 80. The reaction of chlorine gas with solid phosphorus (P 4 produces solid phosphorus pentachloride. hen 16.0 g of chlorine reacts with 3.0 g of P 4, which reactant is limiting? hich reactant is in excess? 10Cl P4(s 0 4PCl 5 (s 16.0 g Cl 1 mol CI 0.6 mol CI 70.90 g CI 1 mol P 4 3.0 g P 4 0.185 moles 13.88 g P 4 According to the balanced equation, Cl reacts with P 4 in a ten-to-one ratio. 1 mol P 4 0.6 mol Cl 10 mol CI 0.06 mol P 4 needed Cl is limiting reactant; P 4 is in excess. 81. Alkaline Battery An alkaline battery produces electrical energy according to this equation. n(s Mn (s (l 0 n(h (s Mn (s a. Determine the limiting reactant if 5.0 g of n and 30.0 g of Mn are used. 1 mol n 5.0 g n 0.380 mol n 65.3 g n 30.0 g Mn 1 mol Mn 86.9 g Mn 0.345 mol Mn According to the balanced equation, Mn reacts with n in a two-to-one ratio. In the reaction, the ratio is 1:1.1 or 0.345/0.380. Mn is the limiting reactant. b. Determine the mass of n(h produced. 0.345 mol Mn 1 mol n(h mol Mn 99.39 g n(h 17.1 g n(h 1 mol n(h 8. Lithium reacts spontaneously with bromine to produce lithium bromide. rite the balanced chemical equation for the reaction. If 5.0 g of lithium and 5.0 g of bromine are present at the beginning of the reaction, determine Li(s Br (l 0 LiBr(s a. the limiting reactant. 1 mol Li 5.0 g Li 3.60 mol Li 6.94 g Li 1 mol Br 5.0 g Br 0.156 mol Br 159.80 g Br Actual ratio of mol Li to mol Br is 3.60 mol Li/0.156 Br or 3:1. nly mol Li are required for 1 mol Br. Thus, Br is the limiting reactant. b. the mass of lithium bromide produced. mol LiBr 0.156 mol Br 0.31 mol LiBr 1 mol Br 86.84 g LiBr 0.31 mol LiBr 7.1 g LiBr 1 mol LiBr c. the excess reactant and the excess mass. Li is in excess. mol Li 0.156 mol Br 0.31 mol Li used 1 mol Br 3.60 mol Li 0.31 mol Li used 3.9 mol Li remaining 6.94 g Li 3.9 mol Li.8 g Li remaining 1 mol Li Section 11.4 Mastering Concepts 83. hat is the difference between actual yield and theoretical yield? Actual yield is the amount of product actually obtained experimentally. Theoretical yield is the amount of product predicted by a stoichiometric calculation. 84. How are actual yield and theoretical yield determined? Actual yield is determined through experimentation. Theoretical yield is calculated from a given reactant or the limiting reactant. Solutions Manual Chemistry: Matter and Change Chapter 11 5
SLUTINS MANUAL 85. Can the percent yield of a chemical reaction be more than 100%? Explain your answer. No, you cannot produce more product than the theoretical yield, which is determined from the starting reactants. 86. hat relationship is used to determine the percent yield of a chemical reaction? actual yeild 100 percent yield theoretical yield 87. hat experimental information do you need in order to calculate both the theoretical and the percent yield of any chemical reaction? The quantity of one reactant and the actual yield of the product 88. A metal oxide reacts with water to produce a metal hydroxide. hat additional information would you need to determine the percent yield of metal hydroxide from this reaction? the mass of one substance in the reaction and the actual mass of metal hydroxide produced 89. Examine the reaction represented in Figure 11.14. Determine if the reaction went to completion. Explain your answer, and calculate the percent yield of the reaction. Element A Element B The reaction did not go to completion. Using squares to represent Element A and circles to represent Element B, the initial products would have yielded four AB particles, but only three were produced. There are enough remaining unreacted A and B particles to produce one more AB particle. The percent yield is 75%. Mastering Problems 90. Ethanol (C H is produced from the fermentation of sucrose (C 1 11 in the presence of enzymes. C 1 11 (aq 0 4C H(l 4C Determine the theoretical yield and the percent yield of ethanol if 684 g of sucrose undergoes fermentation and 349 g of ethanol is obtained. theoretical yield: 1 mol C 684 g C 1 11 1 11 34.3 g C 1 11 4 mol C H 46.07 g C H 1 mol C 1 11 1 mol C H 369 g C H actual yield % yield theoretical yield 100 349 g 100 94.6% 369 g 91. Lead(II oxide is obtained by roasting galena, lead(ii sulfide, in air. The unbalanced equation is: PbS(s 0 Pb(s S a. Balance the equation, and determine the theoretical yield of Pb if 00.0 g of PbS is heated. PbS 3 0 Pb S 1 mol PbS Theoretical yield 00.0 g PbS 39.7 g PbS mol Pb 3.19 g Pb 186.6 g Pb mol PbS 1 mol Pb b. hat is the percent yield if 170.0 g of Pb is obtained? % yield 170.0 g 100 91.10% 186.6 g 6 Chemistry: Matter and Change Chapter 11 Solutions Manual
SLUTINS MANUAL 9. Upon heating, calcium carbonate (CaC decomposes to produce calcium oxide (Ca and carbon dioxide (C. a. Determine the theoretical yield of C if 35.0 g of CaC is heated. CaC (s 0 Ca(s C 1 mol CaC 35 g CaC 100.06 g CaC 1 mol C 43.99 g C 103 g C 1 mol CaC 1 mol C 3 b. hat is the percent yield of C if 97.5 g of C is collected? actual yield % yield theoretical yield 100 97.5 g C 100 94.7% 103 g C 93. Hydrofluoric acid solutions cannot be stored in glass containers because HF reacts readily with silica dioxide in glass to produce hexafluorosilicic acid ( SiF 6. Si (s 6HF(aq 0 SiF 6 (aq (l 40.0 g Si and 40.0 g HF react to yield 45.8 g SiF 6. a. hat is the limiting reactant? 40.0 g Si 1 mol Si 0.666 mol Si 60.09 g Si 1 mol HF 40.0 g HF.00 mol HF 0.01 g HF Ratio of HF to Si in the balanced equation is 6:1. Actual ratio of HF to Si is.00 mol HF or 3.00:1. Thus, HF is the 0.666 mol Si limiting reactant. b. hat is the mass of the excess reactant?.00 mol HF 1 mol Si 6 mol HF 0.333 mol Si 0.666 mol Si available 0.333 mol Si used 0.333 mol Si in excess 0.333 mol Si 60.09 g Si 1 mol Si 0.0 g Si in excess c. hat is the theoretical yield of SiF 6?.00 mol HF 1 mol SiF 6 6 mol HF 0.333 mol SiF 6 0.333 mol SiF 6 144.11 g SiF 6 1 mol SiF 6 48.0 g SiF 6 48.0 g SiF 6 is the theoretical yield. d. hat is the percent yield? % yield 45.8 g SiF 6 100 95.4% yield 48.0 SiF 6 94. Van Arkel Process Pure zirconium is obtained using the two-step Van Arkel process. In the first step, impure zirconium and iodine are heated to produce zirconium iodide (ri 4. In the second step, ri 4 is decomposed to produce pure zirconium. ri 4 (s 0 r(s I Determine the percent yield of zirconium if 45.0 g of ri 4 is decomposed and 5.00 g of pure r is obtained. theoretical yield 45.0 g ri 4 1 mol r 91. g r 6.85 g of r 1 mol ri 4 1 mol r actual yield % yield theoretical yield 100 5.00 g r 100 73.0% 6.85 g r 1 mol ri 4 598.84 g ri 4 Solutions Manual Chemistry: Matter and Change Chapter 11 7
SLUTINS MANUAL 95. Methanol, wood alcohol, is produced when carbon monoxide reacts with hydrogen gas. C 0 CH 3 H hen 8.50 g of carbon monoxide reacts with an excess of hydrogen, 8.5 g of methanol is collected. Complete Table 11.4, and calculate the percent yield for this reaction. According to the balanced equation, Ca 3 (P 4 reacts with Si in a one-to-three ratio. In this reaction, Si is in excess and 0.8060 mol of Ca 3 (P 4 react. Step is to determine amount of P 4 10 produced. 1 mol P 4 10 0.8060 mol Ca 3 (P 4 mol Ca 3 (P 4 0.4030 mol P 4 10 Methanol Reaction Data C CH 3 H(l Mass 8.50 g 9.71 g Molar mass 8.01 g/mol 3.05 g/mol Moles 0.303 mol 0.303 mol 8.50 g C (1 mol C/8.01 g C 0.303 mol C 0.303 mol C (1 mol CH 3 H/1 mol C 0.303 mol CH 3 H 0.304 mol CH 3 H (3.05 g CH 3 H/1 mol CH 3 H 9.71 g CH 3 H Percent yield (8.5 g/9.71 g 100 87.7% yield 96. Phosphorus (P 4 is commercially prepared by heating a mixture of calcium phosphate (CaSi, sand (Si, and coke (C in an electric furnace. The process involves two reactions. Ca 3 (P 4 (s 6Si (s 0 6CaSi (l P 4 10 P 4 10 10C(s 0 P 4 10C The P 4 10 produced in the first reaction reacts with an excess of coke (C in the second reaction. Determine the theoretical yield of P 4 if 50.0 g of Ca 3 (P 4 and 400.0 g of Si are heated. If the actual yield of P 4 is 45.0 g, determine the percent yield of P 4. Step 1 is to determine the excess quantity in equation #1. 400.0 g Si 1 mol Si 6.657 mol of Si 60.08 g Si 1 mol Ca 3 (P 4 50.0 g Ca 3 (P 4 310.17 g Ca 3 (P 4 0.8060 mol Step 3 is to determine amount of P 4 produced in step from 0.4030 mol P 4 10. 1 mol P 4 0.4030 mol P 4 10 13.88 g P 4 1 mol P 4 10 1 mol P 4 49.9 g of P 4 Step 4 is to determine the percent yield, given actual yield is 49.9 g. actual yield % yield theoretical yield 100 45.0 g P 4 100 90.1% 49.9 g P 4 97. Chlorine forms from the reaction of hydrochloric acid with manganese(iv oxide. The balanced equation is: Mn 4HCl 0 MnCl Cl Calculate the theoretical yield and the percent yield of chlorine if 86.0 g of Mn and 50.0 g of HCl react. The actual yield of Cl is 0.0 g. Step 1: The balanced equation is given. Mn (s 4HCl (aq 0 MnCl (aq Cl (l Step is to determine which reactant is in excess. 86.0 g Mn 1 mol Mn 0.989 mol Mn 86.94 g Mn 1 mol HCI 50.0 g HCl 1.37 mol HCI 36.46 g HCI According to the balanced equation, Mn reacts with HCl in a one-to-four ratio. In this reaction, the ratio is 0.989 mol/1.37 mol or 1:1.38. Therefore Mn is in excess, and HCl is the limiting reactant. 8 Chemistry: Matter and Change Chapter 11 Solutions Manual
SLUTINS MANUAL Step 3 is to determine the grams of Cl produced. 1.37 mol HCl 1 mol CI 4 mol HCI 70.90 g CI 1 mol CI 4.3 g CI. Step 4 is to determine percent yield, given that 0.0 g of Cl is actually formed. 0.0 g CI 100 8.3% 4.3 g CI Mixed Review 98. Ammonium sulfide reacts with copper(ii nitrate in a double replacement reaction. hat mole ratio would you use to determine the moles of N N produced if the moles of CuS are known? (N S Cu(N 0 N N CuS mol N N 1 mol CuS 99. Fertilizer The compound calcium cyanamide (CaNCN can be used as a nitrogen source for crops. To obtain this compound, calcium carbide is reacted with nitrogen at high temperatures. CaC (s N 0 CaNCN(s C(s hat mass of CaNCN can be produced if 7.50 mol of CaC reacts with 5.00 mol of N? 1 mol CaNCN 80.11 g CaNCN 5.00 mol N 1 mol N 1 mol CaNCN 401 g CaNCN 100. hen copper(ii oxide is heated in the presence of hydrogen gas, elemental copper and water are produced. hat mass of copper can be obtained if 3.0 g of copper(ii oxide is used? Cu 0 Cu 1 mol Cu 3.0 g Cu 79.55 g Cu 63.55 g Cu 5.6 g Cu 1 mol Cu 1 mol Cu 1 mol Cu 101. Nitrogen oxide is present in urban pollution, but it immediately converts to nitrogen dioxide as it reacts with oxygen. a. rite the balanced chemical equation for the formation of nitrogen dioxide from nitrogen oxide. N 0 N b. hat mole ratio would you use to convert from moles of nitrogen oxide to moles of nitrogen dioxide? mol N mol N 10. Electrolysis Determine the theoretical and percent yield of hydrogen gas if 36.0 g of water undergoes electrolysis to produce hydrogen and oxygen and 3.80 g of hydrogen is collected. (l 0 36.0 g (1 mol /18.0 g.00 mol theoretical yield.00 mol 1 mol mol.0 g 4.04 g H 1 mol % yield 3.80 g 100 94.1% 4.04 g 103. Iron reacts with oxygen as shown. Mass of Fe 30 0 10 0 0 Mass of Fe Formed From Burning Fe 5 10 15 0 5 30 35 Mass of Fe 4Fe(s 1 3 0 Fe (s Different amounts of iron were burned in a fixed amount of oxygen. For each mass of iron burned, the mass of iron(ii oxide formed was plotted on the graph shown in Figure 11.15. hy does the graph level off after 5.0 g of iron is burned? How many moles of oxygen are present in the fixed amount? Solutions Manual Chemistry: Matter and Change Chapter 11 9
SLUTINS MANUAL The graph levels off because the oxygen limits the reaction at that point. 35.0 g Fe 1 mol Fe 159.7 g Fe 0.39 mol Think Critically 3 mole mole Fe 104. Analyze and Conclude In an experiment, you obtain a percent yield of product of 108%. Is such a percent yield possible? Explain. Assuming that your calculation is correct, what reasons might explain such a result? No, percent yields cannot be greater than 100%. High results could mean the product was not completely dry, or it was contaminated. 105. bserve and Infer Determine whether each reaction depends on a limiting reactant. Explain why or why not, and identify the limiting reactant. a. Potassium chlorate decomposes to form potassium chloride and oxygen. No, because there is only one reactant. b. Silver nitrate and hydrochloric acid react to produce silver chloride and nitric acid. Yes, because there are two reactants. Not enough information is given to identify which is the limiting reactant. 106. Design an Experiment Design an experiment that can be used to determine the percent yield of anhydrous copper(ii sulfate when copper(ii sulfate pentahydrate is heated to remove water. btain and record the mass of an empty evaporating dish. Add.00 g of copper(ii sulfate pentahydrate and obtain and record the mass of the hydrate and evaporating dish. Heat the dish gently for 5 minutes, then strongly for 5 minutes to drive off the water. Cool the dish and remass. Record. Determine the mass of the anhydrous copper sulfate. Using the equation CuS 4 5 0 CuS 4 5 and the initial mass of the copper(ii sulfate pentahydrate, determine the theoretical yield of copper(ii sulfate. Determine the actual yield of copper(ii sulfate. Divide the actual yield by the theoretical yield and multiply by 100 to determine percent yield of copper(ii sulfate. 107. Apply Concepts hen a campfire begins to die down and smolder, the flame can be rekindled if you start to fan it. Explain in terms of stoichiometry why the fire begins to flare up again. hen you fan the flame, additional oxygen is added and the remaining coals can burn. 108. Apply Concepts Students conducted a lab to investigate limiting and excess reactants. The students added different volumes of sodium phosphate solution (Na 3 P 4 to a beaker. They then added a constant volume of cobalt(ii nitrate solution (Co(N, stirred the contents, and allowed the beakers to sit overnight. The next day, each beaker had a purple precipitate formed at the bottom. The students decanted the supernatant from each beaker, divided it into two samples, and added one drop of sodium phosphate solution to one sample and one drop of cobalt(ii nitrate solution to the second sample. Their results are shown in Table 11.5. Trial Reaction Data for Co(N and Na 3 P 4 Volume Na 3 P 4 Volume Co(N Reaction with Drop of Na 3 P 4 1 5.0 ml 10.0 ml purple precipitate 10.0 ml 10.0 ml no reaction 3 15.0 ml 10.0 ml no reaction 4 0.0 ml 10.0 ml no reaction Reaction with Drop of Co(N no reaction purple precipitate purple precipitate purple precipitate a. rite a balanced chemical equation for the reaction. Na 3 P 4 (aq 3Co(N (aq 0 Co 3 (P 4 (s 6NaN (aq 30 Chemistry: Matter and Change Chapter 11 Solutions Manual
SLUTINS MANUAL b. Based on the results, identify the limiting reactant and the excess reactant for each trial. Trial 1: Na 3 P 4 is limiting, Co(N is in excess because the addition of Na 3 P 4 caused an additional reaction. Trials -4: Co(N is limiting, Na 3 P 4 is in excess because the addition of Co(N caused and additional reaction. Challenge Problem 109. hen 9.59 g of a certain vanadium oxide is heated in the presence of hydrogen, water and a new oxide of vanadium are formed. This new vanadium oxide has a mass of 8.76 g. hen the second vanadium oxide undergoes additional heating in the presence of hydrogen, 5.38 g of vanadium metal forms. a. Determine the empirical formulas for the two vanadium oxides. 5.38 g V: 0.106 mol 50.94 g/mol 4.1 g : 0.63 mol 15.999 g/mol Divided the smallest molar quantity. 0.106 mol 0.106 mol 1 0.63 mol.48.5 0.106 mol Multiply by the mole ratio by. (1 mol V:.5 mol V 5 5.38 g V: 0.106 mol 50.94 g/mol 3.38 g : 0.11 mol 15.999 g/mol Divided the smallest molar quantity. 0.106 mol 0.106 mol 1 0.11 mol 0.106 mol 1.99 b. rite balanced equations for the steps of the reaction. V 5 0 V V 0 V c. Determine the mass of hydrogen needed to complete the steps of this reaction. 9.59 g V 5 181.88 g/mol 1 mol 1 mol V 5.016 g 0.106 g 1 mol 8.76 g V 8.94 g/mol mol 1 mol V.016 g 0.46 g 1 mol Total hydrogen mass 0.106 g 0.46 g 0.53 g Cumulative Review 110. You observe that sugar dissolves more quickly in hot tea than in iced tea.you state that higher temperatures increase the rate at which sugar dissolves in water. Is this statement a hypothesis or a theory? hy? (Chapter 1 a hypothesis, because it is based only on observation, not on data 111. rite the electron configuration for each of the following atoms. (Chapter 5 a. fluorine [He]s p 5 b. aluminum [Ne]3s 3p 1 c. titanium [Ar]4s 3d d. radon [Xe]6s 4f 14 5d 10 6p 6 1 mol V: mol V Solutions Manual Chemistry: Matter and Change Chapter 11 31
SLUTINS MANUAL 11. Explain why the gaseous nonmetals exist as diatomic molecules, but other gaseous elements exist as single atoms. (Chapter 8 Diatomic molecules achieve a noble gas electron configuration by forming covalent bonds. The monoatomic gases already have noble gas electron configurations. 113. rite a balanced equation for the reaction of potassium with oxygen. (Chapter 9 4K(s 0 K (s 114. hat is the molecular mass of UF 6? hat is the molar mass of UF 6? (Chapter 10 35.0 amu, 35.0 g/mol 38.03 6(18.998 35.0 115. Figure 11.16 gives percent composition data for several organic compounds. (Chapter 10 Percent by mass 50 40 30 0 10 0 5. 13.0 %C %H % 34.8 Percent Composition of Some rganic Compounds 40.0 6.7 53.3 54.5 9.1 36.4 54.5 9.1 36.4 Ethanol Formaldehyde Acetaldehyde Butanoic acid Compound name a. How are the molecular and empirical formulas of acetaldehyde and butanoic acid related? They are whole number multiples of one another. b. hat is the empirical formula of butanoic acid? 54.5 g C (1 mol/1.01 g 4.54 mol 9.1 g H (1 mol/1.01 g 9.0 mol 36.4 g (1 mol/16.00 g.8 mol 4.54 mol C/.8 mol 9.0 mol H/.8 mol 4.8 mol /.8 mol 1 The empirical formula is C. Additional Assessment 116. Air Pollution Research the air pollutants produced by combustion of gasoline in internal combustion engines. Discuss the common pollutants and the reaction that produces them. Show, through the use of stoichiometry, how each pollutant could be reduced if more people used mass transit. Answers will vary. Common pollutants are S, N, N and. Check that stoichiometric calculations account for a decrease in the pollutant. 117. Haber Process The percent yield of ammonia produced when hydrogen and nitrogen are combined under ordinary conditions is extremely small. However, the Haber Process combines the two gases under a set of conditions designed to maximize yield. Research the conditions used in the Haber Process, and find out why the development of the process was of great importance. Answers will vary. Make sure the equation, N 3 NH 3 9 kj, is included. The aim of the Haber process was to control a reaction so that a large amount of a useful product was yielded quickly. The process was of great importance because the Germans had to come up with a nitrogen compound that could be produced in large amounts. 3 Chemistry: Matter and Change Chapter 11 Solutions Manual
SLUTINS MANUAL Document-Based Questions Bombardier Beetles Many insects secrete hydrogen peroxide ( and hydroquinone C 6 (H. Bombardier beetles take this a step further by mixing these chemicals with a catalyst. The result is an exothermic chemical reaction and a spray of hot, irritating chemicals for any would-be predator. Researchers hope to use a similar method to reignite aircraft turbine engines. Figure 11.17 below shows the chemical reaction that results in the bombardier beetle s defensive spray. Data obtained from: Becker, Bob. April 006. ChemMatters. 4: no. H H Energy Catalyst mol C 6 (H from eqauation 4 mol is the limiting reactant. 119. hat is the excess reactant, and how many milligrams are in excess? 1.47 10 3 mol mol C 6 (H 4 mol 110.1 g C 6 (H 1 mg 1 mol C 6 (H 1 10 3 g 80.9 mg C 6 (H used 100.0 mg 80.9 mg 19.1 mg of C 6 (H in excess 10. How many milligrams of benzoquinone will be produced? 1.47 10 3 mol mol C 6 4 mol 108.09 g C 6 1 mg 1 mol C 6 1 10 3 g 79.4 mg C 6 produced C 6 (H Hydroquinone C 6 Benzoquinone 118. If the bombardier beetle stores 100.0 mg of hydroquinone (C 6 (H along with 50.0 mg of hydrogen peroxide (, what is the limiting reactant? Balanced reaction: C 6 (H 4 0 C 6 6 100 mg 50 mg? mg C 6 (H 0 C 6 energy 100.0 mg C 6 (H 1 103 g 1 mg 1 mol 9.08 10 4 mol 110.00 g C 6 (H C 6 (H 50.0 mg 1 103 g 1 mg 1.47 10 3 mol 9.08 10 4 mol C 6 (H 1.47 10 3 mol 1 mol 34.0 g 0.618 mol C 6 (H 1 mol from reaction vs. Standardized Test Practice pages 398 399 1. Stoichiometry is based on the law of a. constant mole ratios. b. Avogadro s constant. c. conservation of energy. d. conservation of mass. d Use the graph below to answer Questions 5. Supply of Various Chemicals in Dr. Raitano s Laboratory AgN 100.0 g NaCl 700.0 g KCl 00.0 g Na P 4 350.0 g Na C 500.0 g Ca(H 300.0 g Solutions Manual Chemistry: Matter and Change Chapter 11 33
SLUTINS MANUAL. Pure silver metal can be made using the reaction shown below. Cu(s AgN (aq 0 Ag(s Cu(N (aq How many grams of copper metal will be needed to use up all of the AgN in Dr. Raitano s laboratory? a. 18.70 g b. 37.3 g c. 74.7 g d. 100 g a 100.0 g AgN 63.55w g Cu 18.70 g Cu 1 mol Cu 1 mol AgN 169.88 g AgN 1mol Cu mol AgN 3. The LeBlanc process is the traditional method of manufacturing sodium hydroxide. The equation for this process is as follows. 4. Pure gas can be generated from the decomposition of potassium chlorate (KCl : KCl (s 0 KCl(s 3 If half of the KCl in the lab is used and 1.8 g of oxygen gas is produced, what is the percent yield of this reaction? a. 1.8% b. 3.7% c. 65.6% d. 98.0% b 1 (00.0 g KCI 100.0 g KCI theoretical yield 100.0 g KCI 1 mol KCI 1.5 g KCI 3 mol 3.00 g 39.17 g mol KCI 1 mol actual yield 1.8 g percent 100 theoretical yield 39.17 g 100 3.7% Na C (aq Ca(H (aq 0 NaH(aq CaC (s Using the amounts of chemicals available in Dr. Raitano s lab, what is the maximum number of moles of NaH that can be produced? a. 4.05 mol b. 4.7 mol c. 8.097 mol d. 9.43 mol c Find the limiting reactant. 1 mol Na C 500.0 g Na C 106.00 g Na C 4.717 mol Na C 300.0 g Ca(H 1 mol Ca (H 74.10 g Ca(H 4.049 mol Ca(H Ca(H is the limiting reactant. mol NaH 4.049 mol Ca(H 1 mol Ca(H 8.097 mol NaH 5. Sodium dihydrogen pyrophosphate (Na P 7, more commonly known as baking powder, is manufactured by heating Na P 4 at high temperatures: Na P 4 (s 0 Na P 7 (s If 444.0 g of Na P 7 is needed, how much more Na P 4 will Dr. Raitano have to buy to make enough Na P 7? a. 0.00 g b. 94.0 g c. 130.0 g d. 480 g c 1 mol Na P 7 444.0 g Na P 7 1.94 g Na P 7 mol Na P 4 119.98 g Na P 4 1 mol Na P 7 1 mol Na P 4 48.0 g Na P 4 480.0 g 350.0 g 130.0 g 34 Chemistry: Matter and Change Chapter 11 Solutions Manual
SLUTINS MANUAL 6. Red mercury(ii oxide decomposes at high temperatures to form mercury metal and oxygen gas: 9. Dimethyl hydrazine (CH 3 N ignites spontaneously upon contact with dinitrogen tetroxide (N 4 : Hg(s 0 Hg(l If 3.55 mol of Hg decomposes to form 1.54 mol of and 618 g of Hg, what is the percent yield of this reaction? a. 13.% b. 4.5% c. 56.6% d. 86.8% d 1 mol theoretical yield 3.55 mol Hg mol Hg 1.775 mol 1.54 mol percent yield 100 86.8% 1.775 mol PERIDIC TABLE 1 Y Y Y Y Y Y Y Y Y Y Y Y Y 3 4 5 6 7 8 9 10 11 1 18 13 14 15 16 17 Y X X X X X X X X X X X X X X X X X X X X X X X X X X X X 7. hich elements tend to have the largest atomic radius in their periods? a. b. X c. Y d. c 8. Elements labeled have their valence electrons in which sublevel? a. s b. p c. d d. f b (CH 3 N (l N 4 (l 0 3N 4 C Because this reaction produces an enormous amount of energy from a small amount of reactants, it was used to drive the rockets on the Lunar Excursion Modules (LEMs of the Apollo space program. If 18.0 mol of dinitrogen tetroxide is consumed in this reaction, how many moles of nitrogen gas will be released? 3 mole N Set up a mole ratio: moles N 4 3 moles N 18 moles N 4 moles N 4 7 moles N Use the table below to answer questions 10 and 11. First Ionization Energy of Period 3 Elements 1st Ionization Element Atomic Number Energy, kj/mol Sodium 11 496 Magnesium 1 736 Aluminum 13 578 Silicon 14 787 Phosphorus 15 101 Selenium 16 1000 Chlorine 17 151 Argon 18 151 Solutions Manual Chemistry: Matter and Change Chapter 11 35
SLUTINS MANUAL 10. Plot the data from this data table. Place atomic numbers on the x-axis. Use the pictures below to answer Questions 13 17. a. Data should form an approximately linear relationship with a few jagged edges, similar to Figure 6.16 on page 191. First Ionization Energy (Kj/mol Atomic Number vs. First Ionization Energy 1600 1400 100 1000 800 600 400 00 10 1 14 16 18 Atomic Number b. c. 11. Summarize the general trend in ionization energy. How does ionization energy relate to the number of valence electrons in an element? d. Ionization generally increases as you move across a period or row in the periodic table. Elements in the first few families have only 1 or valence electrons, which are relatively easy to remove since this will result in a complete outer shell. Elements on the right side of the periodic table have very high ionization energies because their outer shells are nearly filled, therefore making it more likely for these elements to gain a few electrons rather than lose many. 1. How much cobalt(iii titanate (Co Ti 4, in moles, is in 7.13 g of the compound? a..39 10 1 mol b. 3.10 10 mol c. 3. 10 1 mol d. 4.17 10 mol b 7.13 g Co Ti 4 3.10 10 mol 1 mol Co Ti 4 0.0310 mol 9.74 g Co Ti 4 e. 13. Hydrogen sulfide displays this molecular shape. a 14. Molecules with this shape have four shared pairs of electrons and no lone pairs of electrons. c 15. This shape of molecule is referred to as trigonal planar. b 16. Carbon dioxide displays this molecular shape. d 17. Molecules with this shape undergo sp hybridization. b 36 Chemistry: Matter and Change Chapter 11 Solutions Manual