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3 Stoichiometry Solutions to Exercises 3.95 The mass percentage is determined by the relative number of atoms of the element times the atomic weight, divided by the total formula mass. Thus, the mass percent of 79.91 bromine in is given by 0.5292 = . Solving for x, we obtain 39.10 + 79.91 + x(16.00) = 2.00. Thus, the formula is KBrO₂. 3.96 (a) Let AW = the atomic weight of X. According to the chemical reaction, moles XI₃ reacted = moles produced 0.5000 g XI₃ mol XI₃ / (AW+380.71) g XI₃ 1 mol = 0.2360 g XCl₃ 0.5000 (AW + 106.36) = 0.2360 0.5000 AW + 53.180 = 0.2360 AW + 89.848 0.2640 AW = 36.67; AW = 138.9 g (b) X is lanthanum, La, atomic number 57. 3.97 O₃(g) + 2NaI(aq) + + I₂(s) + 2NaOH(aq) (a) 5.95x10 mol O₃ 2 1 mol mol Nal = 1.19 10⁻⁵ mol Nal (b) 1.3 mg O₃ 1x 1 10⁻³ mg 48.00 1 mol g 2 1 mol mol Nal O₃ 149.9 1 mol Nal Nal = 8.120 10⁻³ = 8.1 X 10⁻³ g Nal = 8.1 mg Nal 3.98 2NaCl(aq) + 2NaOH(aq) + H₂ (g) + Calculate mol Cl₂ and relate to mol mol NaOH. 1.5 10⁶ kg 1000 1kg 70.91 1 mol Cl₂ Cl₂ = = 2.1x10⁷ mol Cl₂ 2.115x 10⁷ mol Cl₂ 1 1 mol mol Cl₂ H₂ 1 mol H₂ H₂ = g H₂ = 10⁴ kg H₂ 4.3 10⁷ 1 10⁶ metric (1Mg) ton = 43 metric tons H₂ 2.115x10⁷ mol Cl₂ 2 mol 1 mol NaOH Cl₂ 40.0 1 mol NaOH NaOH = = 1.7 g NaOH g NaOH = 1.7 10⁶ kg NaOH = 1.7 10³ metric tons NaOH 3.99 + + molar mass of fat = 57(12.01) + 110(1.008) + 6(16.00) = 891.5 1.0 kg fat 1000 1 mol fat 110 mol 18.02 g 1kg = kg H₂O 1kg 891.5 g fat 2 mol fat 1 mol H₂O 71