# Question a5357

Aug 14, 2015

You would need 130 g of sodium azide.

#### Explanation:

$\textcolor{red}{2} {\text{NaN"_(3(s)) -> 2"Na"_text((s]) + color(blue)(3)"N}}_{\textrm{2 \left(g\right]}}$

Notice that you have a $\textcolor{red}{2} : \textcolor{b l u e}{3}$ mole ratio between sodium azide and nitrogen gas. This ratio tells you how many moles of nitrogen are produced when a certain number of moles of sodium azide reacts.

SImply pout, regardless of how many moles of sodium azide undergo decomposition, your reaction will produce $\frac{3}{2}$ times more moles of nitrogen gas.

In your case, you have to work backward. You know that the reaction produced enough moles to occupy 75 L at those conditions of pressure and temperature.

Use the ideal gas law equation to determine how many moles of nitrogen were produced

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

n_(N_2) = (99707/101325color(red)(cancel(color(black)("atm"))) * 75color(red)(cancel(color(black)("L"))))/(0.082(color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * (273.15 + 25)color(red)(cancel(color(black)("K")))) = "3.02 moles N"""_2

This means that you must have started with

3.02color(red)(cancel(color(black)("moles N"_2))) * (color(red)(2)" moles NaN"""_3)/(color(blue)(3)color(red)(cancel(color(black)("moles N"_2)))) = "2.013 moles NaN"""_3

Now use sodium azide's molar amss to see how many grams would contain this many moles

2.013color(red)(cancel(color(black)("moles"))) * "65.01 g"/(1color(red)(cancel(color(black)("mole")))) = "130.87 g"#

Rounded to two sig figs, the number of sig figs you gave for the temperature and volume of the gas, the answer will be

${m}_{N a {N}_{3}} = \textcolor{g r e e n}{\text{130 g}}$

A cool video on sodium azide's decomposition