Question

Using 0.082 as your R value, what mass of NaN3 is required to produce 10L of N2 gas, at 273K and 1.5atm. Show all your work completely.Use the following formula to guide your work: 2NaN3 -> 2Na + 3 N2

Answer 1

To solve this question, we need to use the Clapeyron formula.

Clapeyron's formula is a mathematical expression that relates quantities such as pressure (P), volume (V), temperature (T) and the number of particles (n) that make up an ideal gas.

The formula is: PV = nRT

First, let's discover the number in moles of N2. We have:

P = 1.5 atm

V = 10 L

n = ???

R = 0.082 atm.L/mol.k

T = 273 K

1.5 x 10 = n x 0.082 x 273

15 = n x 22.386

n = 15/22.386

n = 0.67 moles

As we can see in the chemical reaction equation, the ratio between NaN3 and N2 is 2:3.

So:

2 mol of NaN3 --- 3 mol of N2

x mol of NaN2 --- 0.67 moles

3x = 2 x 0.67

x = 0.45 moles of NaN3

Now we transform this value into grams, using NaN3 formula mass, which is 65 g/mol:

65 g ---- 1 mol

x g ---- 0.45 mol

x = 29 g

Answer: 29 g of NaN3