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3 Stoichiometry Solutions to Exercises 3.65 (a) + 2A1(OH)₃(s) + 3H₂S(g) (b) Plan. g A mol A mol g B. See Solution 3.61 (c). Solve. 14.2 g Al₂S₃ 150.2 1 mol g 2 1 mol mol 78.00 1 mol = 14.7 g Check. ≈ ≈ 3.66 (a) Ca(OH)₂(aq) + (b) 4.500 g H₂ 2.016 1 mol g H₂ 1 2 mol mol CaH₂ H₂ 42.10 1 mol CaH₂ CaH₂ = 46.99 g CaH₂ 3.67 (a) Analyze. Given: mol Find: mol N₂. Plan. Use mole ratio from balanced equation. Solve. 1.50 mol 3 mol N₂ = 2.25 mol N₂ 2 mol NaN₃ Check. The resulting mol N₂ should be greater than mol (the N₂:NaN₃ ratio is > 1), and it is. (b) Analyze. Given: g N₂ Find: g Plan. Use molar masses to get from and to grams, mol ratio to relate moles of the two substances. Solve. 10.0 g N₂ 28.01 1 mol g N₂ N₂ 2 3 mol mol N₂ 65.01 1 mol NaN₃ NaN₃ = 15.5g NaN₃ Check. Mass relations are less intuitive than mole relations. Estimating the ratio of molar masses is sometimes useful. In this case, 65 g NaN₃/28 g N₂ ≈ 2.25 Then, ≈ 15 g The calculated result looks reasonable. (c) Analyze. Given: vol N₂ in ft³, density N₂ in g/L. Find: g NaN₃. Plan. First determine how many g N₂ are in 10.0 using the density of N₂. Then proceed as in part (b). Solve. 1.25g 1L 1000 1L (2.54)³ 1in³ cm³ X (12)³ 10.0 = 354.0 = 354 g N₂ 354.0 g N₂ 28.01 1 mol N₂ N₂ 2 3 mol mol N₂ 65.01 1 mol NaN₃ = 548 g NaN₃ Check. 1 ~ 28 L; 10 ~ 280 L; 280 L 1.25 ~ 350 g N₂ Using the ratio of molar masses from part (b), (350 2/3 2.25) ≈ 525 g NaN₃ 3.68 (a) 1.50 mol C₈H₁₈ 2 25 mol mol C₈H₁₈ O₂ = 18.75 = 18.8 mol O₂ 60