18 grams of ethylamine was burned in oxygen. Calculate the amount of nitrogen evolved

18 grams of ethylamine was burned in oxygen. Calculate the amount of nitrogen evolved during the complete combustion of this substance.

Let’s write the reaction equation:
4С2H5-NH2 + O2 = 8CO2 + 14H2O + 2N2 ↑
It can be seen from the reaction equation that:
ν (С2H5-NH2) / ν (N2) = 4/2
The amount of a substance can be determined by the expression:
ν = m / M = V / Vm
m (C2H5-NH2) / (4 * M (C2H5-NH2)) = V (N2) / (2 * Vm (N2))
Determine the molar mass of ethylamine:
M (C2H5-NH2) = 12 * 2 + 1 * 5 + 14 + 1 * 2 = 45 g / mol
Let us express the volume of nitrogen from the ratio:
V (N2) = m (C2H5-NH2) * (2 * Vm (N2)) / (4 * M (C2H5-NH2))
Substitute the numbers and find the volume of nitrogen:
V (N2) = 18 * 2 * 22.4 / (4 * 45) = 4.48 liters.
Answer: the volume of nitrogen released during the reaction is 4.48 liters.