How much oxygen is required for the complete combustion of 5 liters of ethylamine?

Given:
V (C2H7N) = 5 l
Find:
V (O2) =?
Solution:
1) Let’s compose the equation of the chemical reaction, we know the volume of ethylamine, and let the volume of oxygen be x l.
4C2H7N + 15O2 = 8CO2 + 14H2O + 2N2
2) Find the molar volume of ethylamine
Vm (C2H7N) = 22.4 L / mol x 4 = 89.6 L / mol
3) Find the molar volume of oxygen
Vm (O2) = 22.4 L / mol x 15 = 336 L / mol
4) 5 L: 89.6 L / mol = x L: 336 L / mol
x = (5 x 336): 89.6 = 18.75 l
Answer: V (O2) = 18.75 liters.