Question

If an automobile air bag has a volume of 11.5 l , what mass of nan3 (in

Answer 1

Moles of NaN3 at STP  = volume of gas / 22.4 = 11.5/22.4 = 0.5mole. Massof NaN3 = moles of NaN3 x molecular weight = 0.5 x 65 = 32.5 grams.

Answer 2

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.