How many grams of sodium azide are needed to provide sufficient nitrogen gas to fill a 50.0 × 50.0 × 25.0 cm bag to a pressure of 1.15 atm at 25.0 °C? 

We're asked to find the mass, in `"g"`, of sodium azide (`"NaN"_3`) needed to produce a certain amount of `"N"_2`.

To do this, we can use the ideal gas equation to find the moles of nitrogen gas present:

`PV = nRT`

• `P = 1.15` `"atm"` (given)

• `V` must be in liters, so we can find the volume first in `"cm"^3` and then convert:

`50.0 xx 50.0 xx 25.0 = 62500` `"cm"^3`

`62500cancel("cm"^3)((1cancel("mL"))/(1cancel("cm"^3)))((1color(white)(l)"L")/(10^3cancel("mL"))) = 62.5` `"L"`

• `R` is the universal gas constant, equal to `0.082057("L"·"atm")/("mol"·"K")`

• `T = 25.0` `""^"o""C"`, which must be in Kelvin:

`T = 25.0` `""^"o""C" + 273 = 298` `"K"`

Plugging in known values, and solving for the number of moles, `n`, we have

`n = (PV)/(RT) = ((1.15cancel("atm"))(62.5cancel("L")))/((0.082057(cancel("L")·cancel("atm"))/("mol"·cancel("K")))(298cancel("K"))) = color(red)(2.94` `color(red)("mol N"_2`

Now, we can use the coefficients of the chemical equation to find the relative number of moles of sodium azide that must react:

`color(red)(2.94)cancel(color(red)("mol N"_2))((2color(white)(l)"mol NaN"_3)/(3cancel("mol N"_2))) = color(green)(1.96` `color(green)("mol NaN"_3`

Finally, we can use the molar mass of sodium azide (`65.01` `"g/mol"`) to find the mass in grams:

`color(green)(1.96)cancel(color(green)("mol NaN"_3))((65.01color(white)(l)"g NaN"_3)/(1cancel("mol NaN"_3))) = color(blue)(ul(127color(white)(l)"g NaN"_3`