# Question 7fb80

Apr 21, 2015

You'd need 52.2 g of sodium azide to produce that much nitrogen gas under those conditions.

$\textcolor{red}{2} N a {N}_{3 \left(s\right)} \to 2 N {a}_{\left(s\right)} + 3 {N}_{2 \left(g\right)}$

Notice that you ahve a $\textcolor{red}{2} : 3$ mole ratio between sidum azide and nitrogen gas; this means that, regardless of how much sodium azide reacts, you'll always produce 3/2 times more nitrogen gas.

Since you already know what your reaction must produce, you can works backwards to determine how much sodium azide must react.

Since you know volume, temperature, and pressure, you can use the ideal gas law equation to determine how many moles of nitrogen gas your reaction must produce

$P V = n R T \implies n = \frac{P V}{R T}$

n_(N_2) = (1.10cancel("atm") * 26.5cancel("L"))/(0.082(cancel("L") * cancel("atm"))/("mol" * cancel("K")) * (273.15 + 22.0)cancel("K")) = "1.204 moles " ${N}_{2}$

According to the aforementioned mole ratio, this many moles of nitrigen would require

1.204cancel("moles "N_2) * (color(red)(2)" moles "NaN_3)/(3cancel("moles "N_2)) = "0.8027 moles "# $N a {N}_{3}$

Now use sodium azide's molar mass to determine how many grams you'd need

$0.8027 \cancel{\text{moles") * "65.01 g"/(1cancel("mole")) = color(green)("52.2 g}}$