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3 Stoichiometry Solutions to Exercises 1440 air x (2.54 cm)³ 0.00118 cm³ air 1000 1kg = 48.12 = 48 kg air 1 air g 48.12 kg air 300 1 mg kg air HCN 1000 1g = 14.43 = 14 HCN mg (b) 2NaCN(s) + Na₂SO₄(aq) + 2HCN(g) The question can be restated as: What mass of NaCN is required to produce 14 g of HCN according to the above reaction? 14.43 g HCNx 1 mol HCN 2 mol NaCN 49.01 NaCN = 26.2 = 26 g NaCN 27.03 g HCN 2 mol HCN 1 mol NaCN (c) 12 ftx 15 ftx 1 X 30 yd² OZ 1lb X 454g = 17,025 16 OZ 1lb = 1.7 10⁴ g acrilan in the room 50% of the carpet burns, so the starting amount of is 0.50(17,025) = 8,513 = 8.5 10³ g 8,513 g CH₂ CHCN 50.9 g HCN = 4333 = g HCN possible 100 g If the actual yield of combustion is 20%, actual g HCN = 4,333(0.20) = 866.6 = 8.7 10² g HCN produced. From part (a), 14 g of HCN is a lethal dose. The fire produces much more than a lethal dose of HCN. 3.112 (a) N₂(g) + O₂(g) 2NO(g); 2NO(g) + O₂(g) 2NO₂(g) (b) 1 million = 1 22 10⁶ tons NO₂ 2000 1 X 453.6 1lb = 1.996 10¹³ = 2.0 10¹³ g ton (c) Plan. Calculate g O₂ needed to burn 500 g octane. This is 85% of total O₂ in the engine. 15% of total O₂ is used to produce according to the second equation in part (a). Solve. + + 500 g C₈H₁₈ 114.2 1 mol g C₈H₁₈ C₈H₁₈ X 2 25 mol mol C₈H₁₈ O₂ X mol O₂ = 1751 = 1.75 10³ 1751 = 0.85; 2060 = 2.1 10³ g O₂ total in engine total g O₂ 2060 g O₂ total 0.15 = 309.1 = 3.1 10² g O₂ used to produce One mol O₂ produces 2 mol NO. Then 2 mol NO react with a second mol O₂ to produce 2 mol Two mol O₂ are required to produce 2 mol one mol O₂ per mol NO₂. 309.1 g O₂ 1 32.00 mol O₂ 1 1 mol mol NO₂ O₂ X 46.01 1 mol O₂ = 444.4 = 4.4 10² g NO₂ 77