Question

What volume of hydrogen will be liberated at STP by the reaction of excess Zn on 7 mL of dil.h2so4 of specific gravity 2.0 and having purity 7 % ? ( h2so4=98g/mol

Answer 1

Given, purity = 7%

specific gravity = 2.0

volume of Dil. H2SO4 = 7ml

molecular weight of H2SO4 = 98 g/mol.

mass of H2SO4 = volume of Dil. H2SO4 × specific gravity × purity

= 7ml × 2g/ml × 7/100

= 98/100 g

so, number of mole of H2SO4 = (98/100)/98 = 0.01mol

we know, reaction between Zn and H2SO4 is ..

$Zn+H_2SO_4\rightarrow H_2+ZnSO_4$

here it is clear that, one mole of H2SO4 liberates one mole of Hydrogen gas.

so, 0.01mole of H2SO4 liberates 0.01mol of Hydrogen gas

so, volume of Hydrogen gas = 0.01 × 22.4 L = 0.224L = 224ml at STP.

Answer 2

Answer:

0.224L

Explanation:

Given, purity = 7%

specific gravity = 2.0

volume of Dil. H2SO4 = 7ml

molecular weight of H2SO4 = 98 g/mol.

mass of H2SO4 = volume of Dil. H2SO4 × specific gravity × purity

= 7ml × 2g/ml × 7/100

= 98/100 g

so, number of mole of H2SO4 = (98/100)/98 = 0.01mol

we know, reaction between Zn and H2SO4 is ..

here it is clear that, one mole of H2SO4 liberates one mole of Hydrogen gas.

so, 0.01mole of H2SO4 liberates 0.01mol of Hydrogen gas

so, volume of Hydrogen gas = 0.01 × 22.4 L = 0.224L = 224ml at STP.