Calculate the amount of oxygen required to burn 20 liters of methylaniline.

Let’s compose an equation and arrange the coefficients:
4СН3NH2 + 9O2 = 4CO2 + 10H2O + 2N2 – reaction of methylamine combustion, carbon monoxide, nitrogen and water are released;
Determine the amount of moles of methylamine:
1 mol of gas at n at – 22.4 l;
X mol (CH3NH2) – 20 liters from here, X mol (CH3NH2) = 1 * 20 / 22.4 = 0.89 mol;
Let’s make the proportion:
0.89 mol (CH3NH2) – X mol (O2);
-4 mol 9 mol hence, X mol (O2) = 0.89 * 9/4 = 2 mol;
We find the volume of oxygen:
V (O2) = 2 * 22.4 = 44.8 liters.
Answer: the reaction will require 44.8 liters of oxygen.