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 by taking one mole of C. Then which option(s) is/are correct?

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

Answer 1

Answer: (B), (D)

The correct options are
B

$\frac{1}{2}$ mole of $A_3{B}_2{C}_2$ is formed.

D

$\frac{1}{2}$ mole of $A_3{B}_2$ is left finally.

$\begin{array}{cccc}& 3A& +2B& \to A_3{B}_2\\ \text{A/c to}& \text{3 moles}& \text{2 moles}& \text{1 mole}\\ \text{stiochilometry}& & & \\ Given:& \text{3 moles}& \text{3 moles}& -\end{array}$

3 moles of A react with 2 moles of B to give 1 mole of $A_3{B}_2$. Hence 1 mole of B would be left unreacted.

$A_3{B}_2+2C\to A_3{B}_2{C}_2\begin{array}{cccc}1mole& 2mole& 1mole& \left(\text{A/c to stoichiometry}\right)\\ 1mole& 1mole& -& \left(\text{Given}\right)\\ \downarrow & & & \\ (\begin{array}{c}\text{from above}\\ \text{reaction}\end{array})& & & \end{array}$

2 moles of C reacts with 1 mole $A_3{B}_2$ to give 1 mole $A_3{B}_2{C}_2$. So, 1 mole of C will react with $\frac{1}{2}$ mole of $A_3{B}_2$ to give $\frac{1}{2}$ mole of $A_3{B}_2{C}_2$. So, $\frac{1}{2}$ mole of $A_3{B}_2$ is left.