Question

The volume of o2 liberated from 0.96g of h2o2 at stp is​

Answer 1

Answer:

2H2O2 → 2H2 + O2

Molecular mass of H2O2 = 68 g/mol

We know that at STP, 1 mole of a gas occupies 22400 ml of volume.

Therefore we can say ,

68 g of H2O2 liberates 22400 ml of O2

Then, 1 g of H2O2 will liberate = 22400/68 ml of O2

∴ The volume of O2 liberated by 0.96 g of H2O2

= 0.96 * 22400/68 ml

= 316.23 ml

Thus, 316.23 ml of O2 is liberated.