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 $C_2H_4$?

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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 $C_2H_4$?

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Md Ariful Islam · Answer 1 · Apr 02, 2024 09:48:30

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.

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 $C_2H_4$?

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