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# chapter4 - Homework Chapter 4 4.4(a 2 Cr(s 3 Cl2(g 2...

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Homework Chapter 4 4.4 (a) 2 Cr(s) + 3 Cl 2 (g) 2 CrCl 3 (s) (b) SiO 2 (s) + 2 C(s) Si(s) + 2 CO(g) (c) 3 Fe(s) + 4 H 2 O(g) Fe 3 O 4 (s) + 4 H 2 (g) 4.10 (a) 454 g Fe 2 O 3 · 1 mol Fe 2 O 3 159.7 g · 2 mol Fe 1 mol Fe 2 O 3 · 55.85 g 1 mol Fe = 318 g Fe (b) 454 g Fe 2 O 3 · 1 mol Fe 2 O 3 159.7 g · 3 mol CO 1 mol Fe 2 O 3 · 28.01 g 1 mol CO = 239 g CO 4.14 (a) BaCl 2 (aq) + 2 AgNO 3 (aq) 2 AgCl(s) + Ba(NO 3 ) 2 (aq) (b) 0.156 g BaCl 2 · 1 mol BaCl 2 208.2 g · 2 mol AgNO 3 1 mol BaCl 2 · 169.9 g 1 mol AgNO 3 = 0.255 g AgNO 3 0.156 g BaCl 2 · 1 mol BaCl 2 208.2 g · 2 mol AgCl 1 mol BaCl 2 · 143.3 g 1 mol AgCl = 0.215 g AgCl 4.22 32.0 g S 8 · 1 mol S 8 256.5 g = 0.125 mol S 8 71.0 g Cl 2 · 1 mol Cl 2 70.91 g = 1.00 mol Cl 2 1.00 mol Cl 2 0.125 mol S 8 = 8.00 mol Cl 2 1 mol S 8 > 4 mol Cl 2 1 mol S 8 S 8 is the limiting reactant Equation S 8 ( l ) + 4 Cl 2 (g) 4 S 2 Cl 2 ( l ) Initial amount (mol) 0.125 1.00 0 Change (mol) –0.125 –0.500 +0.500 Amount after reaction (mol) 0 0.50 0.500 4.26 100. g C 7 H 6 O 3 · 1 mol C 7 H 6 O 3 138.1 g · 1 mol C 9 H 8 O 4 1 mol C 7 H 6 O 3 · 180.2 g 1 mol C 9 H 8 O 4 = 130. g aspirin 100. g C 4 H 6 O 3 · 1 mol C 4 H 6 O 3 102.1 g · 1 mol C 9 H 8 O

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