Question

Ethane$\left(C_2{H}_6\right)$ reacts with oxygen gas to produce carbon dioxide and water. Initially,
5 mol of ethane is burned with 16 mol of oxygen, find the amount of carbon dioxide produced when the yield of the reaction is 60 %.
$2C_2{H}_6+7O_2\to 4C{O}_2+6H_{2}O$

• A. 9.14 mol
• B. 13.25 mol
• C. 5.48 mol
• D. 10.46 mol

Answer 1

The correct option is C 5.48 mol

$2C_2{H}_6+7O_2\to 4C{O}_2+6H_{2}O$
Finding the limiting reagent:
$\text{For Ethane}=\frac{\text{given moles}}{\text{stoichiometric coefficient}}=\frac{5}{2}=2.5\text{For oxygen}=\frac{\text{given moles}}{\text{stoichiometric coefficient}}=\frac{16}{7}=2.29$
Since the ratio is lower for oxygen, it will be the limiting reagent.
According to the reaction:
7 mol of oxygen produces 4 mol of carbon dioxide.
16 mol of oxygen will produce $=\frac{4}{7}\times 16=9.14$ mol of $C{O}_2$ (Theoretical yield)
Since the % yield of the reaction is given as 60 %
% $\text{yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100$
Actual amount of $C{O}_2$ produced = $9.14\times \frac{60}{100}=5.48$ mol