Question

In the reaction $4A+2B+3C \rightarrow A_{4} \, B_{2} \, C_{3},$ what will be the number moles of product formed, starting from one mole of $A,0.6$ moles of $B$ and $0.72$ moles of $C?$

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

Answer 1

Answer: (C)

Stoichiometric mole ratio $ \, \text{A} \, \text{B} \, \, \, \text{C} \\ =\frac{1}{4} \, \, , \, \, \frac{0.6}{2} \, \, , \, \, \frac{0.72}{3} \\ = \, 0.25, \, 0.30,0.24$
$C$ is limiting Reagent
So, according to Stoichiometry of reaction
$3$ mol $C$ gives = $1$ mol $A_{4}B_{2}C_{3}$ .
∴ $0.72$ mol of $C$ gives= $\frac{1}{3} \times 0.72 = 0.24$