Question

An automobile airbag inflates when `NaN_3` is converted to `Na` and `N_2` according to the equation, `2 NaN_3 rarr 2 Na + 3 N_2`. What volume (in L) of `N_2` would be produced if `168" g of NaN"_3` completely reacted at S.T.P.?

Answer 1

`86.8" L of N"_2`

Explanation:

We will need to refer to the equation:

`2 NaN_3 rarr 2 Na + 3 N_2`

We begin Dimensional Analysis by writing the given information as a fraction over 1:

`(168" g of NaN"_3)/1`

Multiply by the conversion factor from grams to moles:

`(168" g of NaN"_3)/1(1" mol of NaN"_3)/(65.0" g of NaN"_3)`

We obtain the conversion factor from moles of `"NaN"_3` to moles of `N_2`, from the equation, and multiply by that factor:

`(168" g of NaN"_3)/1(1" mol of NaN"_3)/(65.0" g of NaN"_3)(3" mol of N"_2)/(2" mol of NaN"_3)`

Multiply by the conversion factor from moles to Liters of a gas at S.T.P.:

`(168" g of NaN"_3)/1(1" mol of NaN"_3)/(65.0" g of NaN"_3)(3" mol of N"_2)/(2" mol of NaN"_3)(22.4" L of N"_2)/(1" mol of N"_2)`

Please take a moment to observe that I have cancelled all of the units, except `"L of N"_2`:

`(168cancel" g of NaN"_3)/1(1cancel" mol of NaN"_3)/(65.0cancel" g of NaN"_3)(3cancel" mol of N"_2)/(2cancel" mol of NaN"_3)(22.4" L of N"_2)/(1cancel" mol of N"_2)`

Perform the multiplication and division:

`168/65 3/2 22.4" L of N"_2 = 86.8" L of N"_2`