What volume of hydrogen gas at STP would be liberated by action of 50 mL of $$H_2SO_4$$ of 50% purity (density = 1.3 g mL$$^{-1}$$) on 20 g of zinc?

Given : Molar mass of H, O, S, Zn are 1, 16, 32, 65 g mol$$^{-1}$$ respectively.

Step 1:

Chemical Equation:

$$Zn+H_2SO_4\longrightarrow\ ZnSO_4+H_2$$

From the stoichiometry, 1 mole of $$Zn$$ reacts with 1 mole of $$H_2SO_4$$ to yield 1 mole of $$H_2$$ gas.

Step 2:

Mass Of Solution:

Volume of acid solution $$=50 mL$$

Density $$=1.3 g mL^(−1)$$

Mass of solution $$=50×1.3=65 g$$

Since acid is $$50\ \%\ $$ pure:

Mass of pure $$H_2SO_4=50/100×65=32.5 g$$

Step 3:

Moles of $$H_2SO_4$$ and $$Zn$$:

Number of Moles $$=$$ $$\ \frac{\ Given\ Mass}{Molar\ Mass}$$

Moles of $$Zn=$$ $$\ \frac{\ 20}{65}\approx\ 0.308\ mol$$

Moles of $$H_2SO_4=$$ $$\ \frac{\ 32.5}{98}\approx\ 0.332\ mol$$

Step 4:

Since, the limiting reagent is $$Zn$$

Therefore, 

Volume of $$H_2$$ at STP:

1 mol $$\approx\ $$ 22.4L

$$V=0.308 mol*22.4 L = 6.892 L$$

Therefore, Volume of $$H_2$$ is $$6.892L$$