# Question 56a51

Dec 3, 2014

The answer is $0.667$.

The combustion of acetylene can be written as

$2 {C}_{2} {H}_{2} + 5 {O}_{2} \to 4 C {O}_{2} + 2 {H}_{2} O$

The balanced chemical reaction shows a $2 : 5$ mole ratio for ${C}_{2} {H}_{2}$ and ${O}_{2}$, a $1 : 2$ ratio for ${C}_{2} {H}_{2}$ and $C {O}_{2}$, and a $1 : 1$ ratio for ${C}_{2} {H}_{2}$ and ${H}_{2} O$.

Now, for simplicity, let us assume that we have $1$ mol of ${C}_{2} {H}_{2}$ to start with ( the equivalence of $26$ grams ); given the fact that the oxygen is in excess, acetylene will always act as a limiting reagent.

So, one would get $2$ moles of $C {O}_{2}$ and $1$ mole of ${H}_{2} O$ in the products, for a total of ${n}_{T} = 2 + 1 = 3$ moles of product.

Therefore, the mole fraction of $C {O}_{2}$ would be

${n}_{C {O}_{2}} = \left({n}_{C {O}_{2}} / {n}_{T}\right) = \frac{2}{3} = 0.667$

In this case, 20%# excess ${O}_{2}$ would mean that you started with $3$ moles of ${O}_{2}$ instead of $2.5$, but I don't see how this would be relevant other than determining that ${O}_{2}$ is not a limiting reagent.