When a car airbag is activated a small quantity of solid sodium azide (NaN₃) reacts very quickly to produce enough nitrogen gas to inflate a nylon bag that cushions a car occupant from injury - Leaving Cert Chemistry - Question c - 2017

To calculate the mass of one mole of sodium azide (NaN₃), we sum the atomic masses of its constituent elements:

• Sodium (Na): 23 g/mol
• Nitrogen (N): 14 g/mol (for three nitrogen atoms: 3 x 14 g/mol)

Thus, the total mass is:

$23 + (3 imes 14) = 23 + 42 = 65 ext{ g/mol}$

To find the number of moles of sodium azide in 130 g, we use the formula:

$n = \frac{m}{M}$

where:

• $m$ = mass of the substance = 130 g
• $M$ = molar mass of sodium azide = 65 g/mol

Substituting the values:

$n = \frac{130}{65} = 2 ext{ moles}$

According to the reaction, for every 2 moles of sodium azide (NaN₃) that decompose, 3 moles of nitrogen gas (N₂) are produced. From the problem, we know that 130 g of NaN₃ produces 46 g of sodium. To determine the mass of nitrogen produced, we first calculate how many moles of sodium were formed:

• Molar mass of sodium = 23 g/mol

So:

$\text{Moles of Na} = \frac{46}{23} = 2 ext{ moles Na}$

Now, since 2 moles of NaN₃ yield 3 moles of N₂, we can find the mass of nitrogen gas produced:

$\text{Moles of N₂ produced} = \frac{3}{2} \times 2 ext{ moles NaN₃} = 3 ext{ moles N₂}$

To find the mass of nitrogen gas produced:

• Molar mass of nitrogen (N₂) = 28 g/mol, Thus: $ext{Mass of N₂} = 3 ext{ moles} \times 28 ext{ g/mol} = 84 ext{ g}$

To find the number of moles of nitrogen (N₂) needed to fill 67.2 litres at standard temperature and pressure (s.t.p), where 1 mole of gas occupies 22.4 litres, we use the formula:

$n = \frac{V}{24.24}$

Substituting the values: $n = \frac{67.2}{22.4} = 3 ext{ moles of N₂}$