Question

In the reaction $4A+2B+3C\to A_4{B}_2{C}_3$
What will be the number of moles of product formed starting from $1\text{mol}$ of $A$, $0.6\text{mol}$ of $B$ and $0.72\text{mol}$ of $C$ ?

• A. $0.25$
• B. $0.3$
• C. $0.24$
• D. $2.32$

Answer 1

The correct option is C $0.24$

In this reaction, $4A+2B+3C\to A_4{B}_2{C}_3$
We have been provided with moles of $A,B$ and $C$. We need to find the limiting reagent here, which is the one whose ratio of actual moles and stoichiometric coefficients is the smallest.
$A:\frac{1}{4}=0.25$

$B:\frac{0.6}{2}=0.3$

$C:\frac{0.72}{3}=0.24$

This makes $C$ the limiting reagent which decides the amount of products formed.
It is clear that $3\text{mol}$ of $C$ produces $1\text{mol}$ of $A_4{B}_2{C}_3$.
By the unitary method, $0.72\text{mol}$ of $C$ produces $0.24\text{mol}$ of it.