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

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i think its 5049.

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