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Chapter 4 Chemical Quantities and Aqueous Reactions 79 Solution: 3 CaCl₂(aq) + 2 Ca₃(PO₄)₂(s) + 6 3 Mg(NO₃)₂(aq) + 2 Mg₃(PO₄)₂(s) + 6 NaCl(aq) 1.5L X 0.050 M CaCl₂ = 0.075 mol CaCl₂ 1.5L X 0.085 M Mg(NO₃)₂ = 0.1275 mol Mg(NO₃)₂ 2 mol 163.94 g mol Na₃PO₄ 0.2025 mol and Mg(NO₃)₂ X 3 mol X = 22 g mol Check: The units of the answer (g Na₃PO₄) are correct. The magnitude of the answer is reasonable because it is needed to remove both the Ca and Mg ions. 4.115 Given: 1.0L; 0.10 M OH⁻ Find: g Ba Conceptual Plan: V,M mol OH mol Ba(OH)₂ mol BaO mol Ba g Ba 1 mol Ba(OH)₂ 1 mol BaO 1 mol Ba 137.33 g Ba 2 mol 1 mol Ba(OH)₂ 1 mol BaO 1 mol Ba Solution: BaO(s) + Ba(OH)₂(aq) 0.10 1 mol 1 1 X 137.33 X g Ba X X L 2 mol OH 1 mol Ba(OH)₂ 1 mol BaO = 6.9 g Ba 1 mol Ba Check: The units of the answer (g Ba) are correct. The magnitude is reasonable because the molar mass of Ba is large and there are 2 moles hydroxide per mole Ba. 4.117 Given: 30.0% NaNO₃, $9.00/100 lb; 20.0% (NH₄)₂SO₄, $8.10/100 lb Find: cost/lb N Conceptual Plan: mass fertilizer mass mass N cost/lb N 30.0 lb 16.48 lb N $9.00 100 lb fertilizer 100 lb 100 lb fertilizer and mass fertilizer mass (NH₄)₂SO₄ mass N cost/lb N 20.0 lb (NH₄)₂SO₂ 21.2 lb N $8.10 100 lb fertilizer 100 lb (NH₄)₂SO₄ 100 lb fertilizer Solution: 30.0 NaNO₃ 16.48 lb N 100 lb fertilizer X X = 4.944 lb N 100 lb fertilizer 100 NaNO₃ $9.00 100 lb fertilizer X 100 lb fertilizer $1.82/lb N 4.944 lb N 20.0 lb 21.2 lb N 100 lb fertilizer X X = 4.240 lb N 100 lb fertilizer 100 $8.10 100 lb fertilizer X 100 lb fertilizer = N 4.24 lb N The more economical fertilizer is the NaNO₃ because it costs less/lb N. Check: The units of the cost ($/lb N) are correct. The answer is reasonable because you compare the cost/lb N directly. 4.119 Given: 24.5 g Au; 24.5 g BrF₃; 24.5 g KF Find: g KAuF₄ Conceptual Plan: g Au mol Au mol 1 mol Au 2 mol 196.97 g Au 2 mol Au g BrF₃ mol BrF₃ mol smaller mol amount determines limiting reactant 1 mol 2 mol KAuF₄ 136.90 g BrF₃ 2 mol BrF₃ g KF mol KF mol 1 mol KF 2 mol 58.10 g KF 2 mol KF then mol KAuF₄ g KAuF₄ 312.07 g KAuF₄ mol Copyright © 2017 Pearson Education, Inc.