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Chapter 12 Practice Set 2. 1. 2 S + 3 O 2  2 SO 3  H  =  791.4 kJ 791.4 kJ 1 mol S 6.44 g S 2 mol S 32.1 g S = 79.4 kJ (79.4 kJ of heat is released.)

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Presentation on theme: "Chapter 12 Practice Set 2. 1. 2 S + 3 O 2  2 SO 3  H  =  791.4 kJ 791.4 kJ 1 mol S 6.44 g S 2 mol S 32.1 g S = 79.4 kJ (79.4 kJ of heat is released.)"— Presentation transcript:

1 Chapter 12 Practice Set 2

2 1. 2 S + 3 O 2  2 SO 3  H  =  791.4 kJ 791.4 kJ 1 mol S 6.44 g S 2 mol S 32.1 g S = 79.4 kJ (79.4 kJ of heat is released.)

3 2. C + O 2  CO 2  H  =  393.5 kJ 393.5 kJ 1 mol C 4.72 g C 1 mol C 12.0 g C = 155 kJ (155 kJ of heat is released.)

4 3. H 2 + Br 2  2 HBr  H  = 72.80 kJ 72.80 kJ 1 mol Br 2 38.2 g Br 2 1 mol Br 2 160. g Br 2 = 17.4 kJ (17.4 kJ of heat is absorbed.)

5 4. 2 P + 5 Cl 2  2 PCl 5  H  =  886 kJ 886 kJ 1 mol Cl 2 1.48 g Cl 2 5 mol Cl 2 71.0 g Cl 2 = 3.69 kJ (3.69 kJ of heat is released.)

6 5. C 2 H 5 OH + 3 O 2  2 CO 2 + 3 H 2 O  H  =  1366.7 kJ 1366.7 kJ 1 mol 4.77 g C 2 H 5 OH C 2 H 5 OH C 2 H 5 OH 1 mol 46.1 g C 2 H 5 OH C 2 H 5 OH = 141 kJ (141 kJ of heat is released.)

7 6. N 2 + O 2  2 NO  H  = 180. kJ 180. kJ 1 mol N 2 13.7 g N 2 1 mol N 2 28.0 g N 2 = 88.1 kJ (88.1 kJ of heat is absorbed.)

8 7. 3 Fe + 2 O 2  Fe 3 O 4  H  =  1120.48 kJ 1120.48 kJ 1 mol Fe 11.8 g Fe 3 mol Fe 55.8 g Fe = 79.0 kJ (79.0 kJ of heat is released.)

9 8. 2 H 2 + O 2  H 2 O  H  =  571.6 kJ 571.6 kJ 1 mol H 2 18.6 g H 2 2 mol H 2 2.02 g H 2 = 2630 kJ (2630 kJ of heat is released.)

10 9. 4 NH 3 + 5 O 2  4 NO + 6 H 2 O  H  =  1170 kJ 1170 kJ 1 mol NH 3 14.9 g NH 3 4 mol NH 3 17.0 g NH 3 = 256 kJ (256 kJ of heat is released...so exothermic.)

11 10. 6 C (graphite) + 3 H 2  C 6 H 6  H  = 49.03 kJ 49.03 kJ 1 mol C 5.81 g C 6 mol C 12.0 g C = 3.96 kJ (3.96 kJ of heat is absorbed...so endothermic.)


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