15. What is the change in enthalpy for the reaction of 4.00 mol of iron(ll) oxid

Question

15. What is the change in enthalpy for the reaction of 4.00 mol of iron(ll) oxide? Show Work 6FeO(s) + Odg)- 2FeOu(s) + 635 kJ a.-635 k b.-423 k c.-106 k d. 106 kJ e. 635 k 16. The enthalpy change for the thermochemical equation is-26.2 kJ. The enthalpy of formation of C Hu(g) in kilojoules per mole must be a.-52.4. b.-26.2 c. 26.2 d. 52.4. e. 104.8. 17. The heat of condensation of water at 373 K is-40.7 kJ/mol. When 9.00 g of water vaporizes at 373 K, the value for AH is Show Work a -366 k. b.-81.4 kJ. c.-20.4 kJ. d. 20.4 kJ. e. 81.4 kJ. 18. Ha for sulfur (S atom s) is 17.7 J How many grams of sulfur can be melted by 22.5 kJ of energy? o Show Work a. 9.8 g b. 19.5 g c. 26.2 g d. 40.7 g e. 52.5 g

Explanation / Answer

15.

6FeO(s) + O2(g) --> 2Fe3O4 (s): H = -635 kJ (exothermic reaction)

H = -635/6 = 105.83 kJ/mol

For 4 moles of FeO(s) : H = -105.83 * 4 kJ = -423.33 kJ

option b.

16.

1/2 C2H4(g) --> C(s) + H2 (g): H = -26.2 kJ

C2H4 formation reaction:

C(s) + H2 (g) --> C2H4 (G)

H = 2 * 26.2 = 52.4 kJ

option d.

17.

H2O (g) --> H2O (l) H = -40.7 kJ/mol

For vaporization:

H2O (l) --> H2O (g) H = 40.7 kJ/mol

Moles of water = 9/18 = 0.5 moles

H = 40.7 * 0.5 = 20.35 kJ

option d.

18.

Hfus = 17.7 kJ/mol

Moles of sulfur that can be melted = 22.5 / 17.7 = 1.27 moles

Mass of sulfur that can be melted = 1.27 * 32 = 40.7 grams

option d.

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