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* 28.3168 = 538.0192 L`

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

`D=(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=(M)/(538.0192 L)`

`1.25 g/L* 538.0192 L= M`

`Mass = 672.524 g`

Now, number of moles of
`N_(2)` gas=
`(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=
`(2 moles of sodium azide)/(3 moles of nitrogen gas)* 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* molar mass of sodium azide`.

=
`16 moles* 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`