Chapter 4 Exercise Key

Transcription

1 Chapter 4 Exercise Key 1 Chapter 4 Exercise Key Exercise 4.1 Balancing Chemical Equations: Balance the following chemical equations. a. P 4 (s) + Cl 4PCl 3 (l) b. 3PbO(s) + NH 3 (g) 3Pb(s) + N (g) + 3H O(l) c. P 4 O 10 (s) + H O(l) 4H 3 PO 4 (aq) d. 3Mn(s) + CrCl 3 (aq) 3MnCl (aq) + Cr(s) e. C H (g) + 5/O (g) CO (g) + H O(l) C H (g) + 5O (g) 4CO (g) + H O(l) f. Fe(NO 3 ) 3 (aq) + 3Na CO 3 (aq) Fe (CO 3 ) 3 (s) + NaNO 3 (aq) g. CH 3 NH (g) + 9/O (g) CO (g) + 5H O(l) + N (g) 4CH 3 NH (g) + 9O (g) 4CO (g) + 10H O(l) + N (g)

2 Chapter 4 Exercise Key Exercise 4. Mass-Mass Equation Stoichiometry: Tetrachloethene, C Cl 4, is a colless liquid used in dry cleaning. Tetrachloethene is often called perchloethylene (perc). It can be fmed in several steps from the reaction of dichloethane, chline gas, and oxygen gas. The equation f the net reaction is below. 8C H 4 Cl (l) + Cl (g) + 7O (g) 4C HCl 3 (l) + 4C Cl 4 (l) + 14H O(l) a. Write fifteen different conversion facts that relate moles of one reactant product to moles of another reactant product. 8 moles C H4Cl 8 moles CH4Cl 8 moles CH4Cl 8 moles CH4Cl 8 moles CH4Cl moles Cl 7 moles O 4 moles CHCl 3 4 moles CCl4 14 moles H O moles Cl moles Cl moles Cl moles Cl 7 moles O 7 moles O 4 moles CHCl3 4 moles CCl4 14 moles HO 4 moles CHCl3 7 moles O 7 moles O 4 moles CHCl 3 4 moles CHCl 3 4 moles CCl4 4 moles CCl4 14 moles HO 4 moles CCl4 14 moles HO 14 moles HO b. What is the maximum mass of perchloethylene, C Cl 4, that can be fmed from 3.75 megagrams of dichloethane, C H 4 Cl? 1 mole C H Cl 4 mole C Cl g C Cl 10 g Mg Mg g CH4Cl 8 mole CH4Cl 1 mole CCl4 10 g? Mg C Cl = 3.75 Mg C H Cl 4 x Mg CCl4? Mg CCl 4 = 3.75 Mg CH4Cl = Mg C Cl 4 8 x Mg CH4Cl c. What is the maximum mass in tons of trichloethene, C HCl 3, fmed with perchloethylene, C Cl 4, in the reaction of 3.75 megagrams of dichloethylene, C H 4 Cl? (Remember that we assume that a ton is an English sht ton unless we are told otherwise.) 10 g 1 mole C H 4Cl 4 mole CHCl g CHCl 3 1 lb 1 ton? tons CHCl 3 = 3.75 Mg CHCl 4 1 Mg g CH4Cl 8 mole CH4Cl 1 mole CHCl g 000 lb 4 x Mg CHCl3 10 g 1 lb 1 ton x Mg CH4Cl 1 Mg 453. g 000 lb? tons C HCl = 3.75 Mg C H Cl = tons CHCl 3

3 Chapter 4 Exercise Key 3 Exercise 4.3 Limiting Reactants: Uranium(IV) oxide, UO, is used as fuel in nuclear power plants. The percentage of the fissionable isotope uranium-35 in the UO used as fuel needs to be higher than that found in nature. To make fuel grade UO, uranium oxides are first converted to uranium hexafluide, UF, which can be enriched in the fissionable isotope of uranium by gas diffusion. The UF is then converted to UO in a series of steps. The chemical equation f the first of these steps is below. UF + H O UO F + 4HF a. What is the maximum mass in pounds of UO F that can be fmed from the reaction of lb UF with 8.0 lb of water? 453. g 1 mole UF 1 mole UO F g UO F 1 lb 1 lb g UF 1 mole UF 1 mole UOF 453. g? lb UO F = lb UF 1 x lb UO F? lb UOF = lb UF 1 x lb UF = 1.47 lb UO F 453. g 1 mole HO 1 mole UOF g UOF 1 lb? lb UOF = 8.0 lb HO 1 lb g HO mole HO 1 mole UOF 453. g 1 x lb UO F? lb UOF = 8.0 lb HO x lb HO b. Why do you think the reactant in excess was chosen to be in excess? = 8 lb UO F Water is much less toxic and less expensive than the radioactive and rare uranium compound. Water in the fm of either liquid steam is also very easy to separate from the solid product mixture.

4 Chapter 4 Exercise Key 4 Exercise 4.4 Equation Stoichiometry and Percent Yield: The raw material f chromium and chromium compounds is a chrome-iron e called chromite. Sodium chromate is made from the roasting of chromite e. Roasting involves heating in the presence of air oxygen. The following equation represents a simplified version of the net reaction. Typically, 75-85% of the chromium in the e can be isolated as sodium chromate. What is the percent yield if 1. tons of Na CrO 4 are produced from e that contains 1.0 ton of FeCr O 4? 4FeCr O 4 + 8Na CO 3 + 7O 8Na CrO 4 + Fe O 3 + 8CO 000 lb 453. g 1 mol FeCrO 4 8 mol Na CrO g Na CrO4 1 lb 1 ton? tons Na CrO 4 = 1.0 ton FeCrO 4 1 ton 1 lb g FeCrO 4 4 mol FeCrO 4 1 mol Na CrO g 000 lb 8 x ton Na CrO 4? tons Na CrO 4 = 1.0 ton FeCrO 4 4 x ton FeCrO 4 actual yield theetical yield 4 Percent Yield = x 100 = x tons Na CrO 1.4 tons Na CrO 4 = 1.4 ton Na CrO 4 = 8% yield

5 Chapter 4 Exercise Key 5 Exercise 4.5 Equation Stoichiometry and Mixtures: The FeS found in pyrite e is a source of sulfur f the production of sulfuric acid. The following equations describe the steps. First the sulfur in FeS is converted to sulfur dioxide, which is converted to sulfur trioxide, which fms sulfuric acid when added to water. 4FeS + 11O Fe O 3 + 8SO V O 5 SO + O Ä SO 3 SO 3 + H O H SO 4 a. Consider the first step in this process. What is the maximum mass of iron(iii) oxide, Fe O 3, fmed when.75 Mg of pyrite e that is 9.% FeS is reacted with excess oxygen in air? 9. Mg FeS 10 g 1 mole FeS mole Fe O g Fe O3 1 Mg? Mg FeO 3 =.75 Mg e 100 Mg e 1 Mg g FeS 4 mole FeS 1 mole FeO3 10 g 9. Mg FeS x Mg FeO3? Mg FeO 3 =.75 Mg e 100 Mg e 4 x Mg FeS = 1.9 Mg Fe O 3 b. What is the minimum mass of pyrite e that is 9.% FeS necessary to produce 1.00 Mg of iron(iii) oxide, Fe O 3?? Mg e = 1.00 Mg Fe O 1 mole Fe O 4 mole FeS g FeS 10 g 3 1 Mg 100 Mg e 3 1 Mg g FeO3 mole FeO3 1 mole FeS 10 g 9. Mg FeS? Mg e = 1.00 Mg Fe O 4 x g FeS 100 Mg e 3 = 1. Mg e x Mg FeO3 9. Mg FeS

6 Chapter 4 Exercise Key Exercise 4. Writing Combustion Reactions: Write balanced equations f the complete combustion of (a) C 4 H 10 (g), (b) C 3 H 7 OH(l), and (c) C 4 H 9 SH(l). a. C 4 H 10 (g) + 13O (g) 8CO (g) + 10H O(l) b. C 3 H 7 OH(l) + 9O (g) CO (g) + 8H O(l) c. C 4 H 9 SH(l) + 15O (g) 8CO (g) + 10H O(l) + SO (g) Exercise Classification of Chemical Reactions: Classify each of these reactions as a combination reaction, a decomposition reaction, a combustion reaction, a single-displacement reaction. a. HgO(s) Hg(l) + O (g) decomposition b. C 1 H O 11 (s) + 1O (g) 1CO (g) + 11H O(l) combustion c. B O 3 (s) + 3Mg(s) B(s) + 3MgO(s) single-displacement d. C H 4 (g) + H (g) C H (g) combination