What volume of nitrogen gas at STP is produced when 68.2 g of trinitrotoluene, C7H5(NO2)3 reacts?

10.07 L is the volume for the produced N₂

Explanation:

This is a question for the Ideal Gases Equation:

P . V = n . R . T

In STP conditions, the pressure is 1 atm and the T° is 273K

Let's determine the moles of produced N₂  by the decomposition of TNT

The reaction is:

2 C₇H₅(NO₂)₃ (s) → 7 C (s) + 7 CO (g) + 3 N₂ (g) + 5 H₂O (g)

We convert the mass to moles (mass / molar mass)

68.2 g . 1 mol / 227.13 g = 0.300 moles

Ratio is 2:3. 2 moles of TNT can produce 3 moles of N₂

Then, 0.300 moles will produce (0.300 . 3) / 2 = 0.450 moles

We replace data:

1 atm . V = 0.450 mol . 0.082 L.atm/mol.K . 273K

V = (0.450 mol . 0.082 L.atm/mol.K . 273K) / 1 atm

V = 10.07 L

4d^55s^1 ( third choice)