Question

What amount of $NaN{O}_3$ (in grams) that has a percentage purity of 30% produces 0.56 litre of $O_2$ at STP (atm) on decomposition?
Given reaction: $2NaN{O}_3\to 2NaN{O}_2+O_2$

• A. 18.42 g
• B. 14.16 g
• C. 21.59 g
• D. 7.13 g

Answer 1

The correct option is B 14.16 g

As per the reaction: $2NaN{O}_3\to 2NaN{O}_2+O_2$

22.4 L of $O_2$ is produced by decomposition of 2 moles of $NaN{O}_3$
0.56 L of $O_2$ is produced by $\frac{2}{22.4}\times 0.56=0.05$ moles of $NaN{O}_3$
Amount of pure $NaN{O}_3$ decomposed $=0.05mol\times 85$ (g/mol) = 4.25 g
But purity of sample = 30% (given)
Actual amount of $NaN{O}_3$ required $=\frac{4.25}{30}\times 100=14.16$ g