Chemistry 65 ANSWER KEY REVIEW QUESTIONS Chapter 8 1. Given the unbalanced equation shown below: NaClO (s) NaCl (s) + O (g) a) How many grams of O can be produced from reaction of 1.0 moles of NaClO? mol O.00 g 1.0 mol NaClO x x = 576 g O mol NaClO b) How many grams of NaCl are produced when 80.0 g of O are produced? mol NaCl 58.5 g 80.0 g O x x x = 97. g NaCl.00 g mol O. Given the balanced equations below, how many moles of each reactant must react to produce 1.70 mole of N? NH (g) + 6 NO (g) 5 N (g) + 6 HO (g) mol NH 1.70 mol N x = 10.96 mol NH 5 mol N 6 mol NO 1.70 mol N x = 16. mol NO 5 mol N 1
. What mass of CO can be produced from the reaction of 5.0 g of CH8 with 75.0 g of O according to the following equation: CH8 + 5 O CO + HO mol CO 5.0 g C H x x = 1.70 mol CO 8.11 g CH8 mol CO 75.0 g O x x = 1.06 mol CO Limiting Reactant.00 g 5 mol O.01 g 1.06 mol CO x = 61.9 g CO. How many grams of SO are produced when 15 g of CS react with 8.0 g of O according to the following equation: CS + O CO + SO mol SO 15 g CS x x =.99 mol SO 76.1 g CS mol SO 8.0 g O x x = 1.00 mol SO Limiting Reactant.00 g mol O 6.06 g 1.00 mol CO x = 6.1 g SO 5. When 50.0 g of MgCO react completely with HPO, 15.8 g of CO are produced. What is the percent yield for this reaction? The unbalanced equation is given below: HPO + MgCO Mg(PO) + CO + HO mol CO.01 g 50.0 g MgCO x x x = 6.1 g CO 8.1 g mol MgCO 15.8 g % Yield = x100 = 60.5% 6.1 g
6. The reaction of calcium and nitrogen gas (shown below) can be carried out with 90.0% yield. In an experiment, 56.6 g of calcium are reacted with 0.5 g of nitrogen gas. What mass of calcium nitride is formed in this experiment? Ca (s) + N (g) CaN (s) Ca N 18.6 g 0.08 g mol Ca 56.6 g Ca x x x = 69.8 g CaN Limiting Reactant Ca N 18.6 g 0.5 g N x x x = 161 g Ca N 8.0 g N 90.0 g 100 g 69.8 g CaN x = 6.8 g CaN 7. Hydrogen gas has been suggested as a clean fuel because it produces only water when it burns. If the reaction has a 98% yield, what mass of hydrogen gas would form 75.0 kg of water? H (g) + O (g) HO (g) Actual yield 75.0 kg Theoretical yield = = = 76.5 kg HO % yield 0.98 mol H.0 g 76.5 kg H O x x x = 8.58 kg H 18.0 g mol HO 8. Diborane (BH6) can be produced by the unbalanced reaction shown below. If the reaction has an 85.0% yield, how many grams of NaBH are needed to produce 0.0 g of diborane? NaBH (s) + BF (g) BH6 (g) + NaBF (s) Actual yield 0.0 g Theoretical yield = = =.5 g B H % yield 0.850 6 mol NaBH 7.8 g.5 g B H x x x = 8. g NaBH 6 7.68 g mol BH6
9. Consider the reaction shown below: BH6 (g) + 6 Cl (g) BCl (g) + 6 HCl (g) H= 755 kj a) Is this reaction endothermic or exothermic? Exothermic b) How much heat is released when 85.0 g of BH6 react? O 755 kj 85.0 g BH 6 x x = 0 kj ( sig figs) 7.68 g BH6 10. Thermal decomposition of mercury (II) oxide occurs as shown below: HgO (s) Hg (l) + O (g) H = 181.6 kj a) How much heat is needed to decompose 555 g of HgO? 181.6 kj 555 g HgO x x = kj ( sig figs) 16.59 g mol HgO b) If 75 kj of heat is absorbed, how many grams of mercury form? mol Hg 00.59 g 75 kj x x = 608 g Hg ( sig figs) 181.6 kj
11. Iron (II) sulfide reacts with HCl according to the reaction: FeS (s) + HCl (aq) FeCl (aq) + HS (g) A reaction mixture initially contains 0.0 g of FeS and.8 g of HCl. When the reaction has occurred as completely as possible, what mass of which reactant is left? mol HCl 6.6 g 0.0 g FeS x x x = 16.6 g HCl 87.91 g FeS FeS 87.91 g.8 g HCl x x x = 8.7 g FeS 6.6 g mol HCl Calculations above show that FeS is limiting reactant since 8.7 g FeS is required to react with all of the HCl (.8 g) and only 0.0 g is available. Therefore 7. g of HCl is in excess after the reaction is complete (.8 g 16.6 g). 1. People often use sodium bicarbonate as an antacid to neutralize excess hydrochloric acid in an upset stomach. (a) Write a balanced equation for the reaction of aqueous sodium bicarbonate and aqueous hydrochloric acid; (b) What mass of hydrochloric acid (in grams) can.5 g of sodium bicarbonate neutralize? NaHCO (aq) + HCl (aq) NaCl (aq) + CO (g) + HO (l) HCl 6.6 g.5 g NaHCO x x x = 1.1 g HCl 8.01 g NaHCO 1. A solution contains an unknown mass of dissolved barium ions. When sodium sulfate is added to the solution, a white precipitate forms. The precipitate is filtered, dried and found to have a mass of 58 mg. Based on this information, what mass of barium was in the original sample? (Assume that all of the barium was precipitated out of the solution by the reaction.) Ba + + NaSO (aq) BaSO (s) + Na + + 1 g Ba 17. g 58 mg BaSO x x x x = 0.15 g 10 mg.9 g BaSO 5
1. Consider the reaction shown below: NH (g) + NO (g) N (g) + HO (g) A reaction flask initially contains 7.5 g of NH and 7.9 g of NO. Determine the identity and mass of all the substances present in the reaction flask after all the reactants have reacted as much as possible. (Assume 100% yield.) N O 9.0 g 7.5 g N H x x x = 9.5 g N O reacted.06 g mol NH Since amount of N O reacted is less than amount available (7.9 g), it is in excess. Therefore, at end of reaction: excess N O = 7.9 g -9.5 g = 5. g mol N 8.0 g 7.5 g N H x x x = 6.1 g N.06 g mol NH mol H O 18.0 g.06 g mol N H 7.5 g NH x x x = 0.9 g HO 15. Magnesium ions can be precipitated from seawater by addition of sodium hydroxide. How many grams of NaOH must be added to a sample of seawater to completely precipitate 88. mg of magnesium present? (Write a balanced equation for the reaction first.) Mg + NaOH (aq) Mg(OH) (s) + Na + 1 g mol NaOH 0.00 g 10 mg.1 g Mg + 88. mg Mg x x x x = 0.91 g NaOH + 16. Pyrite (FeS), an impurity in some coals, reacts with oxygen to form the air pollutant sulfur dioxide, as shown below. What mass of SO (in grams) is produced when.0x10 kg of coal containing 0.050 mass % pyrite is burned? FeS (s) + SO (g) SO (g) + FeO (s) 10 g 0.050 g FeS mol SO 6.06 g.0x10 kg coal x x x x = 1.1x10 g SO 1 kg 100 g coal 119.97 g mol FeS 6