Answer:
0.641 moles of ethane
Explanation:
Based on the equation:
C2H6(g) + 7/2O2(g) → 2CO2(g) + 3H2O(l)
We can determine ΔH of reaction using Hess's law. For this equation:
Hess's law: ΔH products - ΔH reactants
ΔH = {2ΔHCO2 + 3ΔHH2O} - {ΔHC2H6}
Pure monoatomic substances have a ΔH = 0kJ/mol; ΔHO2 = 0kJ/mol
ΔH = {2*-393.5kJ/mol + 3*-285.8kJ/mol} - {-84.7kJ/mol}
ΔH = -1559.7kJ/mol
That means when 1 mole of ethane is in combustion there are released 1559.7kJ of heat. To produce 1.00x10³kJ there are needed:
1.00x10³kJ * (1mole ethane / 1559.7kJ) =