Hydrogen sulfide burns form sulfur dioxide:

2h2s(g) +3o2(g) → 2so2(g) + 2h2o(g) δh= -1036 kj

calculate the enthalpy change when burning 26.7 g of hydrogen sulfide in kj

pls

404.04 kJ.

Explanation:

To calculate the moles, we use the equation:

moles of

According to stoichiometry :

2 moles of on burning produces = 1036 kJ

Thus 0.78 moles of on burning produces =

Thus the enthalpy change when burning 26.7 g of hydrogen sulfide is 404.04 kJ.

i believe the answer is a.

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