Question

Use the reaction below for the decomposition of sodium azide

Answer 1

Answer: The volume of nitrogen gas at STP is 44.8 L.

Explanation:

The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation for it follows:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$                    ...(1)

We are given:

Given mass of $NaN_3$ = 130.0 g

Molar mass of $NaN_3$ = 65.01 g/mol

Using equation 1:

$Moles of NaN_3=\frac{130.0 g}{65.01g/mol}=2mol$

For the given chemical equation:

$2NaN_3(s) \rightarrow 2Na(s) + 3N_2(g)$

By the stoichiometry of the reaction:

2 moles of $NaN_3$ produces 3 moles of nitrogen gas

At STP:

1 mole of a gas occupies 22.4 L of volume

So, 2 moles of nitrogen gas will occupy $=\frac{22.4L}{1mol}\times 2mol=44.8L$ of volume

Hence, the volume of nitrogen gas at STP is 44.8 L.