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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.
Solution
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
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