Question

Use the reaction below for the decomposition of sodium azide

to sodium metal and nitrogen gas.

2NaNg(s) + 2Na(s) + 3N2(g)

What volume of nitrogen at STP is generated by the decomposition of 130.0 g NaNz?

Answer 1

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

Step-by-step 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=(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
`=(22.4L)/(1mol)* 2mol=44.8L` of volume

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