Question

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`?

Answer 1

There is insufficient dioxygen to combust the ethylene completely.

Explanation:

We represent the combustion of ethylene by this reaction:

`H_2C=CH_2(g) +3O_2(g) rarr 2CO_2(g) + 2H_2O(l)`

For complete combustion, `28*g` of ethylene thus requires `96*g` of dioxygen.

Here, `"moles of ethylene"` `=` `(1.60xx10^2*g)/(28.05*g*mol^-1)=5.70*mol`.

And, this quantity of ethylene requires, `3xx5.70*molxx32.00*g*mol^-1` `"dioxygen gas"` `=` `547.6*g`, approx. `17*mol`.

However, we can certainly represent incomplete combustion:

`H_2C=CH_2(g) +2O_2(g) rarr 2CO(g) + 2H_2O(l)`;

And `H_2C=CH_2(g) +O_2(g) rarr 2C(s) + 2H_2O(l)`.

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