#100 The combustion of 1 mole of octane fuel corresponds to the following reaction equation. C_{8}H_{8} (l) + 25/2 O_{2}(g) --> 8 CO_{2} (g) + 9 H_{2}O (g) You will also need the molar mass of C_{8}H_{8} = 114.2 g/mol, and CO_{2} = 44.01 g/mol. Volume of octane used = 125.0 mi / [19.50 mi/gal] = 6.410 gal 6.410 gal X [3.7854 L/gal] = 24.27 L --> 24.27 X 10^{3} mL Mass of octane used = 24.27 X 10^{3} mL X [0.690 g/mL] = 1.674 X 10^{4} g Mol of octane used = 1.674 X 10^{4} g / [114.2 g/mol] = 1.466 X 10^{2} mol From the reaction equation: 1 mol of octane generates 8 mol of CO_{2} Mass of CO_{2} generated : [5.575 X 10^{2} mol X 8] X [44.01 g/mol] = 5.162 X 10^{4} g --> 51.62 kg #101 The ratio Ag:AgNO_{3} is 0.634985 This means 0.634985 = [Molar mass of Ag] / [Molar mass of AgNO_{3}] Molar mass of Ag = 107.8682 g/mol Molar mass of O = 15.9994 g/mol Molar mass of NO_{3} = Ψ + (3 X 15.9994 g/mol) [107.8682 g/mol] / [107.8682 g/mol + Ψ + (3 X 15.9994 g/mol)]= 0.634985 = [Ag] / [Ag + N + (3 X O)] [107.8682 g/mol] = 0.634985 X [107.8682 g/mol + Ψ + (3 X 15.9994 g/mol)] [107.8682 g/mol] = 68.4947 g/mol + 0.634985Ψ + 30.4781 g/mol [107.8682 g/mol] = 98.9728 g/mol + 0.634985Ψ [107.8682 g/mol] - 98.9728 g/mol = 0.634985Ψ [8.8954 g/mol] = 0.634985Ψ [8.8954 g/mol] / 0.634985 = Ψ --> Ψ = 14.0088 g/mol Accepted value = 14.0067 g/mol Absolute error = 14.0088 g/mol - 14.0067 g/mol = +0.0021 g/mol Relative error = [+0.0021 g/mol / 14.0067 g/mol] X 100 = 0.015 % #102a Chemical equations: Burning sulfur Reaction with calcium oxide #102b 1 ton = 2000 lb, 1 lb = 2200 g Coal usage per day = 2000 tons X 2000 lb/ton = 4.000 X 10^{6} lb 4.000 X 10^{6} lb / 2.20 lb/kg = 1.82 X 10^{6} kg = 1.82 X 10^{9} g Since this coal is 2.5 % sulfur by mass, Grams of sulfur burnt per day = 1.82 X 10^{9} g X 0.0250 = 4.545 X 10^{7} g Mol of sulfur burnt per day = 4.545 X 10^{7} g / [32.065 g/mol] = 1.418 X 10^{6} mol From the chemical equations... one mol of sulfur generates one mole of calcium sulfite. Using the molar mass of calcium sulfite = 120 g/mol. Grams of calcium sulfite produced per day = 1.418 X 10^{6} mol X [120 g/mol] = 1.701 X 10^{8} g |