If 0.5 g of a mixture of two metals A and B with respective equivalent weights 12 and 9 displace 560 mL of H₂ at STP from an acid, the composition of the mixture is

According to STP conditions, one mole of a gas has a volume of 22400 mL. At STP conditions, one mole of H₂ has a volume of 22400 mL. 

If one mole of H₂ (or 2 equivalents of H₂) has a volume of 22400 mL, then one equivalent of H₂ has a volume of 11200 mL.

If one equivalent of H₂ has a volume of 11200 mL, then the number of equivalents of H₂ present in 560 mL is calculated as:

$\begin{array}{ccc}11200 \mathrm{mL} \mathrm{of} {\mathrm{H}}_2& =& \mathrm{One} \mathrm{equivalent} \mathrm{of} {\mathrm{H}}_2\\ 560 \mathrm{mL} \mathrm{of} {\mathrm{H}}_2& =& 560 \mathrm{mL} \mathrm{of} {\mathrm{H}}_2\times \frac{\mathrm{One} \mathrm{equivalent} \mathrm{of} {\mathrm{H}}_2}{11200 \mathrm{mL} \mathrm{of} {\mathrm{H}}_2}\\ & =& \frac{1}{20} \mathrm{equivalents} \mathrm{of} {\mathrm{H}}_2\end{array}$

Total mass of the mixture is 5.6 g. Assume that the mass of A is x g and the mass of B is 0.5-x g.

Equivalent mass of A is calculated as:

$\begin{array}{ccc}\mathrm{Number} \mathrm{of} \mathrm{equivalents} \mathrm{of} \mathrm{A}& =& \frac{\mathrm{Given} \mathrm{mass}}{\mathrm{Equivalent} \mathrm{mass} \mathrm{of} \mathrm{A}}\\ & =& \frac{\mathrm{x}}{12}\end{array}$

Equivalent mass of B is calculated as:

$\begin{array}{ccc}\mathrm{Number} \mathrm{of} \mathrm{equivalents} \mathrm{of} \mathrm{B}& =& \frac{\mathrm{Given} \mathrm{mass}}{\mathrm{Equivalent} \mathrm{mass} \mathrm{of} \mathrm{B}}\\ & =& \frac{0.5-\mathrm{x}}{9}\end{array}$

Equivalents of A and B displaces the equivalents of H₂. This implies that sum of equivalents of A and B is equal to the equivalents of H₂ present in 560 mL.

Substitute the values in the above equation and determine the value of x.

$\begin{array}{ccc}\frac{\mathrm{x}}{12}+\frac{0.5-\mathrm{x}}{9}& =& \frac{1}{20}\\ \frac{3\mathrm{x}+4(0.5-\mathrm{x})}{36}& =& \frac{1}{20}\\ \mathrm{x}& =& 0.2\end{array}$

Therefore, mass of A is 0.2 g and the mass of B is 0.3 g (0.5 g-0.2 g).

Composition of A in the mixture is calculated as:

$\begin{array}{ccc}\mathrm{Composition} \mathrm{of} \mathrm{A} \mathrm{in} \mathrm{mixture}& =& \frac{\mathrm{Mass} \mathrm{of} \mathrm{A}}{\mathrm{Total} \mathrm{mass} \mathrm{of} \mathrm{the} \mathrm{mixture}}\times 100\\ & =& \frac{0.2}{0.5}\times 100\\ & =& 40\%\end{array}$

Composition of B in the mixture is calculated as:

$\begin{array}{ccc}\mathrm{Composition} \mathrm{of} \mathrm{B} \mathrm{in} \mathrm{mixture}& =& \frac{\mathrm{Mass} \mathrm{of} \mathrm{B}}{\mathrm{Total} \mathrm{mass} \mathrm{of} \mathrm{the} \mathrm{mixture}}\times 100\\ & =& \frac{0.3}{0.5}\times 100\\ & =& 60\%\end{array}$