You are given 4.5 moles of O_2 to react with 1.60 x10^2 g C_2H_4. Upon completion of the reaction, will there be any remaining C_2H_4?

Nov 16, 2016

There is insufficient dioxygen to combust the ethylene completely.

Explanation:

We represent the combustion of ethylene by this reaction:

${H}_{2} C = C {H}_{2} \left(g\right) + 3 {O}_{2} \left(g\right) \rightarrow 2 C {O}_{2} \left(g\right) + 2 {H}_{2} O \left(l\right)$

For complete combustion, $28 \cdot g$ of ethylene thus requires $96 \cdot g$ of dioxygen.

Here, $\text{moles of ethylene}$ $=$ $\frac{1.60 \times {10}^{2} \cdot g}{28.05 \cdot g \cdot m o {l}^{-} 1} = 5.70 \cdot m o l$.

And, this quantity of ethylene requires, $3 \times 5.70 \cdot m o l \times 32.00 \cdot g \cdot m o {l}^{-} 1$ $\text{dioxygen gas}$ $=$ $547.6 \cdot g$, approx. $17 \cdot m o l$.

However, we can certainly represent incomplete combustion:

${H}_{2} C = C {H}_{2} \left(g\right) + 2 {O}_{2} \left(g\right) \rightarrow 2 C O \left(g\right) + 2 {H}_{2} O \left(l\right)$;

And ${H}_{2} C = C {H}_{2} \left(g\right) + {O}_{2} \left(g\right) \rightarrow 2 C \left(s\right) + 2 {H}_{2} O \left(l\right)$.

Both of these reactions would occur to some extent in any combustion reaction. Given the limited quantity of oxidant, some ethylene would remain unreacted.