Use the information below to answer this question
c(s) + o2(g) → co2(g) ∆h = −393.5 kj mol−1
h2(g) + o2(g) → h2o(l) ∆h = −285.8 kj mol−1
3c(s) + 4h2(g) → c3h8(g) ∆h = −104.0 kj mol−1
4c(s) + 5h2(g) → c4h10(g) ∆h = −125.2 kj mol−1
the value in kj mol−1
for the enthalpy of combustion of propane is

First, construct a balance equation for the combustion of propane. Then the 3rd equation given in the question is the formation of propane. So 1 mol of propane is formed by -104 kJ mol-1. Then, the 1st equation given in question is the formation of CO2. 3 mol of CO2 is formed so 3(-393.5). 2nd equation shows that 1 mol of H2O is formed but based on balanced equation that I constructed, 4 mol of H2O is formed so 4(-285.5). The arrows I drew was for the mathematics equation. Going to same direction of the product of combustion of propane FROM the reactant is -104 and x so I put this on the left side. Then, the remaining one is on the right side.