chemistry equations

Question

The reaction for the combustion of propane is

C3H8(g)+5O2(g)⟶3CO2(g)+4H2O(l)Δ𝐻=−526.3kcalmol

How much heat is released when 31.4 g of propane is burned?

Answer 1

C₃H₈(g) + 5O₂(g) ⟶ 3CO₂(g) + 4H₂O(l) ... Δ𝐻 = −526.3kcalmol

molar mass C₃H₈ = 44 g/mol

31.4 g C₃H₈ x 1mol / 44 g = 0.714 moles C₃H₈

0.714 moles x 526.3 kcal/mol = 376 kcal of heat released

You can write this as -376 kcal