Question

At 25.0◦C and 1.10 atm, what volume of N2(g) is produced by the decomposition of 71.4 g NaN3 under the following chemical reaction: 2NaN3(s) → 2Na(l) + 3N2(g)

Answer 1

Solution :

Balanced reaction equation

  2NaN3(s) → 2Na(l) + 3N2(g)

Lets first calculate the moles of the N2 that can be produced from 71.4 NaN3

Moles = mass/ moles

(71.4 g NaN3* 1 mol / 65.01 g) *(3 mol N2/2 mol NaN3) = 1.6474 mol N2

So it can produce 1.6474 mol N2

Now lets calculate the volume of the N2 gas at 25 C and 1.10 atm

25 C +273 = 298 K

PV= nRT

V= nRT / P

V= 1.6474 mol * 0.08206 L atm per mol K * 298 K / 1.10 atm

V= 36.62 L

So the volume of the N2 gas = 36.62 L