Question

question 14 of 24
balance the following combustion reaction in order to answer the question. use lowest whole - number coefficients.
combustion reaction: $c_2h_4 + o_2\
ightarrow co_2 + h_2o$
a combustion reaction occurs between 9.5 mol $o_2$ and 143 g $c_2h_4$. upon completion of the reaction, is there any $c_2h_4$ remaining?
no, all of the $c_2h_4$ is used up in the reaction.
yes, there is still $c_2h_4$ remaining.

Explanation:

Step1: Balance the combustion reaction

First, balance carbon atoms. There are 2 carbon atoms in $C_2H_4$, so we need 2 moles of $CO_2$. Then balance hydrogen atoms. Since there are 4 hydrogen atoms in $C_2H_4$, we need 2 moles of $H_2O$. Finally, balance oxygen atoms. On the right - hand side, we have $2\times2 + 2\times1=6$ oxygen atoms, so we need 3 moles of $O_2$ on the left - hand side. The balanced reaction is $C_2H_4+3O_2
ightarrow2CO_2 + 2H_2O$.

Step2: Calculate the moles of $C_2H_4$

The molar mass of $C_2H_4$ is $M=(2\times12 + 4\times1)\text{ g/mol}=28\text{ g/mol}$. Given the mass of $C_2H_4$ is $m = 143\text{ g}$, the number of moles of $C_2H_4$, $n_{C_2H_4}=\frac{m}{M}=\frac{143\text{ g}}{28\text{ g/mol}}\approx5.11\text{ mol}$.

Step3: Determine the limiting reactant

From the balanced equation, the mole ratio of $C_2H_4$ to $O_2$ is $1:3$. For $5.11\text{ mol}$ of $C_2H_4$, the moles of $O_2$ required is $n_{O_2\text{(required)}}=3\times5.11\text{ mol}=15.33\text{ mol}$. But we have only $9.5\text{ mol}$ of $O_2$. So $O_2$ is the limiting reactant.

Step4: Check if $C_2H_4$ remains

Since $O_2$ is the limiting reactant, $C_2H_4$ is in excess. So there will be $C_2H_4$ remaining.

Answer:

Yes, there is still $C_2H_4$ remaining.