C4H10(g) + 13/2O2(g) => 4CO2(g) + 5H2O(g)+ Q
4×♢fH°CO2(g) + 5×♢fH°H2O(g) - ♢fH°C4H10(g)+13/2×♢fH°O2(g) = 4×(-393,2) + 5×(-241,6) - (-126,03) +0 = -1572,8 - 1208 + 126,03 = -2654,77 KJ/mol
1 mol C4H10...........................2654,77 KJ
x moli C4H10..........................10619,08 KJ
x = 4 moli C4H10