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?

1 Answer
anor277
Nov 16, 2016

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.

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