Question

$A+B\to A_3{B}_2$ (unbalanced)
$A_3{B}_2+C\to A_3{B}_2{C}_2$ (unbalanced)
Above two reactions are carried out by taking 3 moles each of $A$ and $B$ and one mole of $C$. Then which option is/are correct?

• A. 1 mole of $A_3{B}_2{C}_2$ formed
• B. 1/2 mole of $A_3{B}_2{C}_2$ formed
• C. 1/2 mole of $A_3{B}_2$ formed
• D. 1/2 mole of $A_3{B}_2$ is left finally

Answer 1

Answer: (B)

1/2 mole of $A_3{B}_2{C}_2$ formed

$A+B\to A_3{B}_2$ (unbalanced)
$3A+2B\to A_3{B}_2$ (balanced)
$A_3{B}_2+C\to A_3{B}_2{C}_2$ (unbalanced)
$A_3{B}_2+2C\to A_3{B}_2{C}_2$ (balanced)
The overall reaction will be
$3A+2B+2C\to A_3{B}_2{C}_2$
According to above balanced chemical equations, the mole ratio
$A:B:C=3:2:2$
Thus, 2 moles of C will react with 2 moles of B and 3 moles of A.
Thus, 1 mole of C will react with 1 mole of B and 1.5 moles of A. But 3 moles of A and 3 moles of B are present. Hence, A and B are excess reactants and C is limiting reactant.
2 moles of C gives 1 mole of $A_3{B}_2{C}_2$.

1 mole of C will give $\frac{1}{2}$ mole of $A_3{B}_2{C}_2$.

Note:
3 moles of A and 3 moles of B will react to give 3 moles of $A_3{B}_2$.
2 moles of C will combine with 1 mole of $A_3{B}_2$.
1 mole of C will combine with $\frac{1}{2}$ mole of $A_3{B}_2$.
$3-\frac{1}{2}=\frac{5}{2}$ moles of $A_3{B}_2$ will be left out.