Can someone help me with an enthalpy change question?

Question

Hydrogen sulfide, H2S, is a foul-smelling gas. It burns to form sulfur dioxide.

2H2S(g) + 3O2(g) → 2SO2(g) + 2H2O(g) ∆H = −1037. kJ

Calculate the enthalpy change to burn 57.19 g of hydrogen sulfide.

___ kJ

Answer 1

*Standard* Enthalpies of formation (25C)
in kcal/mole
H2S: -4.82
O2: 0
So2: -70.96
H2O: --57.8
Products: 2*(-71)+2*(-58)=-258
Reactants 2*(-4.8)=-9.6
Total: 248 kcal/mol,

grams /mol of H2S-:
34

moles H2S in problem 57/34 = 1.68

248 kcal/mol * 1.68/2 = 208

208 "kilogram Calories" * 4.19 kJ/kcal = 872 kJ

Check: 1.68 * 1037 / 2 = 871

May be called the "heat of stoichiometric combustion"