Question

$3A+2B\to A_3{B}_2$
$A_3{B}_2+2C\to A_3{B}_2{C}_2$
The above two reactions are carried out by taking 2 mol of each of B and C and 3 mol of A. Then which option(s) is/are correct?

• A. 1 mol of $A_3{B}_2{C}_2$ is formed
• B. Half mol of $A_3{B}_2{C}_2$ is formed
• C. Half mol of $A_3{B}_2$ is formed in first reaction
• D. 1 mol of $A_3{B}_2$ is formed in first reaction

Answer 1

Answer: (A), (D)

1 mol of $A_3{B}_2$ is formed in first reaction

$3A+2B\to A_3{B}_2\text{(balanced)}$
$A_3{B}_2+2C\to A_3{B}_2{C}_2\text{(balanced)}$

Now, 3 mol of A reacts with 2 mol of B to form 1 mol of $A_3{B}_2$.
So, 1 mol of $A_3{B}_2$ is formed when 3 mol of A and 2 mol of B are taken.
2 mol of C reacts with 1 mol of $A_3{B}_2$ to give 1 mol of $A_3{B}_2{C}_2$.
Therefore, correct options are (a) and (d).