# 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?

Jul 29, 2017

$127$ ${\text{g NaN}}_{3}$

#### Explanation:

We're asked to find the mass, in $\text{g}$, of sodium azide (${\text{NaN}}_{3}$) needed to produce a certain amount of ${\text{N}}_{2}$.

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

$P V = n R T$

• $P = 1.15$ $\text{atm}$ (given)

• $V$ must be in liters, so we can find the volume first in ${\text{cm}}^{3}$ and then convert:

$50.0 \times 50.0 \times 25.0 = 62500$ ${\text{cm}}^{3}$

62500cancel("cm"^3)((1cancel("mL"))/(1cancel("cm"^3)))((1color(white)(l)"L")/(10^3cancel("mL"))) = 62.5 $\text{L}$

• $R$ is the universal gas constant, equal to $0.082057 \left(\text{L"·"atm")/("mol"·"K}\right)$

• $T = 25.0$ $\text{^"o""C}$, which must be in Kelvin:

$T = 25.0$ $\text{^"o""C} + 273 = 298$ $\text{K}$

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 ($65.01$ $\text{g/mol}$) 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