**Step 1. Write the balanced chemical equation.**

The balanced equation is

#"2NaN"_3 → "2Na" + "3N"_2#

**Step 2. Strategy**

The problem is to convert grams of #"NaN"_3# to moles of #"N"_2# and volume of #"N"_2#.

We can use the flow chart below to help us.

(Adapted from www.lsua.us)

The process is:

**(a)** Use the **molar mass** to convert the mass of #"NaN"_3# to moles of #"NaN"_3#.

**(b)** Use the **molar ratio** (from the balanced equation) to convert moles of #"NaN"_3# to moles of #"N"_2#.

**(e)** Use the **Ideal Gas Law** to convert moles of #"N"_2# to volume of #"N"_2#.

In equation form,

#"grams of NaN"_3 stackrelcolor(blue)("molar mass"color(white)(ml)) (→) "moles of NaN"_3 stackrelcolor(blue)("molar ratio"color(white)(ml))→ "moles of N"_2 stackrelcolor(blue)("Ideal Gas Law"color(white)(ml))(→) "volume of N"_2#

**The Calculations**

**(a) Moles of #"NaN"_3#**

#156 color(red)(cancel(color(black)("g NaN"_3))) × ("1 mol NaN"_3)/( 65.01 color(red)(cancel(color(black)("g NaN"_3)))) = "2.40 mol NaN"_3 #

**(b) Moles of #"N"_2#**

#2.40color(red)(cancel(color(black)("mol NaN"_3))) × ("3 mol N"_2)/(2 color(red)(cancel(color(black)("mol NaN"_3)))) = "3.60 mol N"_2#

**(c) Volume of #"N"_2#**

The Ideal Gas Law is

#color(blue)(bar(ul(|color(white)(a/a)PV = nRTcolor(white)(a/a)|)))" "#

We can rearrange this to give

#V = (nRT)/P#

#n = "3.60 mol"#

#R = "8.314 Pa·m"^3·"K"^"-1""mol"^"-1"#

#T = "(27 + 273.15) K" = "300.15 K"#

#P = 1.1 × 10^5color(white)(l) "Pa"#

∴ #V = (3.60 color(red)(cancel(color(black)("mol"))) × 8.314 color(red)(cancel(color(black)( "Pa")))·"m"^3·color(red)(cancel(color(black)("K"^"-1""mol"^"-1"))) × 300.15color(red)(cancel(color(black)( "K"))))/(1.1 × 10^5 color(red)(cancel(color(black)("Pa")))) = "0.082 m"^3 = "82 dm"^3#

The volume of #"N"_2# produced is #"82 dm"^3#.