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Mixture of two bivalent metals $A$ and $B$ having mass $2 \mathrm{~g}$ when dissolved in $\mathrm{HCl}$ at STP, $2.24 \mathrm{~L}$ $\mathrm{H}_2$ is evolved. What is the mass of $A$ present in mixture? (Atomic mass of $A=15 \mathrm{u}, B=30 \mathrm{u}$ )
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The correct answer is:
$1 \mathrm{~g}$
$\quad A+2 \mathrm{HCl} \rightarrow A \mathrm{Cl}_2+\mathrm{H}_2$
Mole : $\frac{x}{15}$
$\frac{x}{15}$
$B+2 \mathrm{HCl} \rightarrow \mathrm{BCl}_2+\mathrm{H}_2$
Mole : $\frac{2-x}{30}$
$\frac{2-x}{30}$
Total moles of $\mathrm{H}_2=\frac{x}{15}+\frac{2-x}{30}=\frac{2.24}{22.4}=\frac{1}{10}$ $\Rightarrow \quad x=1 \mathrm{~g}$
Mole : $\frac{x}{15}$
$\frac{x}{15}$
$B+2 \mathrm{HCl} \rightarrow \mathrm{BCl}_2+\mathrm{H}_2$
Mole : $\frac{2-x}{30}$
$\frac{2-x}{30}$
Total moles of $\mathrm{H}_2=\frac{x}{15}+\frac{2-x}{30}=\frac{2.24}{22.4}=\frac{1}{10}$ $\Rightarrow \quad x=1 \mathrm{~g}$
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