Step 1. Write the balanced chemical equation.
The balanced equation is
"2NaN"_3 → "2Na" + "3N"_22NaN3→2Na+3N2
Step 2. Strategy
The problem is to convert grams of "NaN"_3NaN3 to moles of "N"_2N2 and volume of "N"_2N2.
We can use the flow chart below to help us.
![Flow Chart]()
(Adapted from www.lsua.us)
The process is:
(a) Use the molar mass to convert the mass of "NaN"_3NaN3 to moles of "NaN"_3NaN3.
(b) Use the molar ratio (from the balanced equation) to convert moles of "NaN"_3NaN3 to moles of "N"_2N2.
(e) Use the Ideal Gas Law to convert moles of "N"_2N2 to volume of "N"_2N2.
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"_2grams of NaN3molar massml−−−−−−−→moles of NaN3molar ratioml−−−−−−−→moles of N2Ideal Gas Lawml−−−−−−−−−→volume of N2
The Calculations
(a) Moles of "NaN"_3NaN3
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.
