Question

In the reaction
$4A+2B+3C\to A_4{B}_2{C}_3$
The number of moles of product formed, starting from one mole of A, 0.6 mole of B and 0.72 mole of C?

• A. 0.25
• B. 0.3
• C. 0.24
• D. 0.32

Answer 1

The correct option is C 0.24

Mole ratio according to the stoichiometry
$A:B:C$ = $4:3:2$
Given mole ratio:
$A:B:C$ = $1:0.6:0.72$
The limiting reagent would be the one that has a smaller ratio of available moles to the stoichiometric coefficients.
$A:B:C=\frac{1}{4}:\frac{0.6}{2}:\frac{0.72}{3}=0.25:0.3:0.24$
Hence C, is the limiting reagent.
Moles of product that can be formed from 0.72 mole of C = $\frac{1}{3}\times 0.72=0.24$