Question

When 1 gram of a mixture of aliminium and zinc was treated with HCl, a gas was liberated. At the end of the reaction, the volume of the liberated gas was found to be $524c{m}^3$, under STP conditions. The individual weights of aluminium and zinc in the mixture, respectively, are:

• A. 0.2 g and 0.8 g
• B. 0.8 g and 0.2 g
• C. 0.5 g and 0.5 g
• D. 0.322 g and 0.678 g

Answer 1

The correct option is A 0.2 g and 0.8 g

Let mass of $Al$ in mixture be $x$ $grams$

$\therefore$ mass of $Zn$ in mix will be $1-x$ $grams$

Moles of $Al=\frac{x}{27}$ $moles$

Moles of $Zn=\frac{1-x}{65}$ $moles$

$Al+3HCl⟶AlC{l}_3+\frac{3}{2}{H}_2$

$Zn+2HCl⟶ZnC{l}_2+H_2$

$\therefore$ Moles of $H_2$ from $Al=\frac{3}{2}\times \frac{x}{27}=\frac{x}{18}$ $moles$

Moles of $H_2$ from $Zn=1\times \frac{1-x}{65}=\frac{1-x}{65}$ $moles$

Volume of liberated $H_2=524$ $c{m}^3$ at STP

$1$ $mole$ gas at STP occupies $22400$ $c{m}^3$ volume

$\therefore$ Moles of $H_2$ liberated $=\frac{524}{22400}$ $moles$

$\therefore \frac{524}{22400}=\frac{x}{18}+\frac{1-x}{65}$

$⟹x=0.2$ $g$

$\therefore$ Weight of aluminium $=0.2$ $g$

Weight of Zinc $=0.8$ $g$