How much heat is produced by burning 4.00 moles of acetylene under standard state conditions?

Combustion of acetylene is represented as:

C2H2 (g) + 5/2O2 (g) → 2CO2 (g) + H2O (l)

ΔHrxn = ∑nΔH°f (products) - ∑nΔH°f (reactants)

n = # moles

ΔH°f = standard enthalpy of formation from thermochemical tables

ΔHrxn = [2ΔH°f (CO2) + ΔH°f (H2O) ] - [ΔH°f (C2H2) + 3/2ΔH°f (O2) ]

= [2 (-393.5) - 285.8] - [226.7 + 0] = - 1299.5 kJ/mol

Based on the above:

1 mole of acetylene produces - 1299.5 kJ of heat

Therefore, heat produced by burning 4 moles is = - 1299.5 * 4 = - 5198 kJ

Ans: - 5198 kJ of heat is produced by burning 4.00 moles of acetylene under standard state conditions