What will be the change in enthalpy when 100.0 g of butane, C4H10, is burned in oxygen as shown in the thermochemical equation below?

2 C4H10(l) + 13 O2(g) → 8 CO2(g) + 10 H2O(g) ΔH = −5271 kJ

−2636 kJ
−4534 kJ
−9087 kJ
−1.817 × 104 kJ

3 answers

First find how many moles is 100.0g of butane. Divide 100.0g / 58.14.
So 1.72 mol of butane=-5271J.
The mol coefficient of butane C4H10 is 2.
So multiply the mole of butane by the coefficient to get -1.817*104kj

Do.... Your.... Own..... Work.

4534Kj
(5271/2)*1.72