Question

A mixture of two bivalent metals weighing $2g(M_A=15,M_B=30)$ is dissolved in $HCl$ to evolve $2.24L{H}_2$ at $STP$. What is the mass of metal $A$ present in the mixture?

• A. $1g$
• B. $1.5g$
• C. $0.5g$
• D. $0.75g$

Answer 1

The correct option is A $1g$

Let the mass of A in the mixture be $x$.
Since metal A is bivalent, it reacts with $HCl$ as shown below:
$A+2HCl\to AC{l}_2+H_2$
From the above chemical equation,
$1\text{mol}$ of $A$ liberates $1\text{mol}$ of $H_2$.
$\therefore$ $\frac{x}{15}\text{mol}$ of $A$ liberates $=\frac{x}{15}\text{mol}$ of $H_2$
Thus, Volume of $H_2$ liberated at STP $=\text{moles}\times \text{molar volume}=\frac{x\times 22.4}{15}L$

As metal $B$ is also bivalent, the second reaction takes place as shown below:
$B+2HCL\to BC{l}_2+H_2$
From the above chemical equation,
$1\text{mol}$ of $B$ liberates $1\text{mol}$ of $H_2$.
$\therefore$ $\frac{2-x}{30}\text{mol}$ of $B$ liberates $=\frac{2-x}{30}\text{mol}$ of $H_2$
Thus, Volume of $H_2$ liberated at STP $=\text{moles}\times \text{molar volume}=\frac{(2-x)\times 22.4}{30}L$
Therefore, total mole of $H_2$ liberated$=\frac{x}{15}+\frac{2-x}{30}=\frac{2.24}{22.4}=\frac{1}{10}$
$⟹\frac{x}{15}-\frac{x}{30}=\frac{1}{10}-\frac{1}{15}$
$⟹x=1$
Thus mass of $A$ in the mixture is $1g$.
Hence the correct answer is option (a).