Question

How many grams of nan3 are required to produce 19.0 ft3 of nitrogen gas, about the size of an automotive air bag, if the gas has a density of 1.25 g/l?

Answer 1

The  balanced chemical reaction is given as:

$2NaN_{3}(s)\rightarrow 2Na(s)+3N_{2}(g)$

Now, convert $19.0 ft^{3}$ into litres.

$1 ft^{3} = 28.3168$

So, $19.0 ft^{3} = 19\times 28.3168 = 538.0192 L$

Density is equal to the ratio of mass to the volume.

$D=\frac{M}{V}$

where, M = mass and V= volume $(538.0192 L)$

Substitute the value of density and volume in formula to get the value of mass.

$1.25 g/L=\frac{M}{538.0192 L}$

$1.25 g/L\times 538.0192 L= M$

$Mass = 672.524 g$

Now, number of moles of $N_{2}$ gas=$\frac{672.524 g}{28.02 g/mol}$

= $24.00 moles$

According to the reaction, 2 moles of sodium azide gives 3 moles of nitrogen gas.

Now, in 24.00 moles of nitrogen gas produced from= $\frac{2 moles of sodium azide}{3 moles of nitrogen gas}\times 24.00 moles$ of nitrogen gas, moles of sodium azide.

number of moles of sodium azide  = $16 moles$

Mass of sodium azide in g  =  $number of moles\times molar mass of sodium azide$.

= $16 moles\times 65.00 g/mol$

= $1040 g$

Thus, mass of sodium azide which is required to produce $19.0 ft^{3}$ of nitrogen gas  = $1040 g$