# Question #42736

Mar 26, 2017

Let's assume that $\text{100 mol}$ of hydrocarbon are combusted:

#### Explanation:

$30 {C}_{3} {H}_{8} + 70 {C}_{4} {H}_{10} + 605 {O}_{2} \rightarrow 370 C {O}_{2} + 470 {H}_{2} O$

And of course I could divide this by 10 to get prime coefficients:

$3 {C}_{3} {H}_{8} + 7 {C}_{4} {H}_{10} + \frac{121}{2} {O}_{2} \rightarrow 37 C {O}_{2} + 47 {H}_{2} O$

Which is (?) balanced with respect to mass and charge. Is it?

So you need a tad over 6 equiv of dioxygen gas per 10 equiv hydrocarbon mix to combust the hydrocarbon completely. Anyway, come back for another serve if you are unsatisfied; you have not asked a question yet.