Question

If an automobile air bag has a volume of 11.5 l , what mass of nan3 (in
g. is required to fully inflate the air bag upon impact? assume stp conditions.

Answer 1

Answer: The mass of sodium nitride required is 35.4 grams.

Explanation:

We are given:

Volume of gas = 11.5 L

At STP:

22.4 L of volume is occupied by 1 mole of gas

So, 11.5 L of volume will be occupied by = $\frac{1}{22.4}\times 11.5=0.513mol$ of sodium nitride

To calculate the mass for given number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Molar mass of sodium nitride = 69 g/mol

Moles of sodium nitride = 0.513 moles

Putting values in above equation, we get:

$0.513mol=\frac{\text{Mass of sodium nitride}}{69g/mol}\\\\\text{Mass of sodium nitride}=(0.513mol\times 69g/mol)=35.4g$

Hence, the mass of sodium nitride required is 35.4 grams.

Answer 2

22.3 g of NaN₃ are required to fully inflate an airbag of 11.5 L at STP.

In airbags, sodium azide decomposes to form sodium and nitrogen gas, which inflates the bag. The decomposition reaction is:

2 NaN₃ ⇒ 2 Na + 3 N₂

We can calculate the mass of NaN₃ needed to produce 11.5 L of N₂ at STP, using the following relations.

• At STP, 1 mole of N₂ occupies 22.4 L.
• The molar ratio of N₂ to NaN₃ is 3:2.
• The molar mass of NaN₃ is 65.01 g/mol.

$11.5 L N_2 \times \frac{1molN_2}{22.4LN_2} \times \frac{2molNaN_3}{3molN_2} \times \frac{65.01 gNaN_3}{1molNaN_3} = 22.3 g NaN_3$

22.3 g of NaN₃ are required to fully inflate an airbag of 11.5 L at STP.

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