# Question #a5357

##### 1 Answer

You would need **130 g** of sodium azide.

#### Explanation:

Start with the balanced chemical equation for this decomposition reaction

#color(red)(2)"NaN"_(3(s)) -> 2"Na"_text((s]) + color(blue)(3)"N"_text(2(g])#

Notice that you have a

SImply pout, regardless of how many moles of sodium azide undergo decomposition, your reaction will produce

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

#PV = nRT implies n = (PV)/(RT)#

#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_(NaN_3) = color(green)("130 g")#

A cool video on sodium azide's decomposition