What volume of hydrogen at N.T.P would be liberated by the action of 50 mL of dilute `H_(2)SO_(4)` of 40 % purity and having a specific gravity of `1.3 g mL^(-1)` on 65 g of zinc ? (Atomic mass of Zn = 65) ?

Answer

5.94 L

Explanation

Mass of 50 mL of dilute `H_(2)SO_(4)` = volume `xx` density
`=(50 mL )xx ("1.3 g mL"^(-1))=65g`
Mass of pure `H_(2)SO_(4)` in the sample `= ((65g)xx40)/(100)=26g`
The balanced chemical equation is :
`Zn+underset(underset(98g)("1 mol"))(H_(2)SO_(4))rarrZnSO_(4)+underset(underset(22.4L)("1 mol"))(H_(2))`
98 g of `H_(2)SO_(4)` evolve `H_(2)` at N.T.P = 22.4 L
26 g `H_(2)SO_(4)` evolve `H_(2)` at N.T.P `= ((22.4L))/((98.0g))xx(26.0g)=5.94L`.