Question

The volume of $O_2$ obtained at 2 atm and 546 K, by the complete decomposition of 8.5 g $NaN{O}_3$ is $2NaN{O}_3\to 2NaN{O}_2+O_2$

• A. 2.24 lit
• B. 1.12 lit
• C. 0.84 lit
• D. 0.56 lit

Answer 1

The correct option is A 2.24 lit

Molar mass of $NaN{O}_3=23+14+3\times \left(16\right)=85g⟹\frac{1}{10}$ mole of $NaN{O}_3.$

$2$ moles of $NaN{O}_3$ produces $1$ mole of $O_2,$

Let, $\frac{1}{10}$ mole of $NaN{O}_3$ produces $x$ moles of $O_2.$

$x\times 2=\frac{1}{10}\times 1x=\frac{1}{20}$

$1$ mole of $O_2⟹32gms⟹22.4Liters$ (Avagadro's law).

$\frac{1}{20}$ molews of $O_2⟹\frac{32}{20}gms⟹\frac{22.4}{20}Liters$

Volume of $O_2$ obtained is $1.12$ Liters.