Question

2NaN3 -> 2Na + 3N2 Determine the volume of nitrogen gas generated at STP from the decomposition of 120.5g NaN3. [Molar mass: Na = 22.99; N = 14]

Answer 1

The balanced chemical equation for the decomposition of $\sf{NaN_3}$ is,

$\sf{2NaN_3_{(s)}\longrightarrow 2Na_{(s)}+3N_2_{(g)}}$

From this equation we get that 2 moles of $\sf{NaN_3}$ is decomposed to form 3 moles of $\sf{N_2}$ gas.

$\longrightarrow\sf{2\ mol\ NaN_3\equiv3\ mol\ N_2}$

Mass of 1 mole of $\sf{NaN_3=22.99+3\times14=64.99\ g.}$

Volume of 1 mole of $\sf{N_2}$ gas at STP $\sf{=22.4\ L.}$

Hence,

$\longrightarrow\sf{2\times 64.99\ g\ NaN_3\equiv3\times 22.4\ L\ N_2}$

$\longrightarrow\sf{129.98\ g\ NaN_3\equiv67.2\ L\ N_2}$

$\longrightarrow\sf{129.98\times\dfrac{120.5}{129.98}\ g\ NaN_3\equiv67.2\times\dfrac{120.5}{129.98}\ L\ N_2}$

$\longrightarrow\underline{\underline{\sf{120.5\ g\ NaN_3\equiv62.3\ L\ N_2}}}$

Therefore, $\sf{120.5\ g\ NaN_3}$ gets decomposed to form $\bf{62.3\ L}\ \sf{N_2}$ gas at STP.