Question

1.56 g of sodium peroxide reacts with water according to the following equation:

\[\ce{2Na2O2 + 2H2O -> 4NaOH + O2}\]

Calculate the volume of oxygen liberated at STP.

Answer 1

156 g of Na₂O₂ liberates oxygen at S.T.P. = 22.4 L

∴ 1.56 g of Na₂O₂ liberates oxygen

= `22.4/156 xx 1.56`

= 0.224 L