Determine the amount of oxygen required to burn 100 grams of ethylamine.

Given:
m (C2H5NH2) = 100 g
To find:
V (O2) =?
Decision:
1) Let’s compose the equation of the chemical reaction, the mass of ethylamine is known to us, and let the volume of oxygen be x l.
4С2H5NH2 + 15O2 = 8CO2 + 14H2O + 2N2
2) Find the molecular weight of ethylamine
M (C2H5NH2) = 12 x 2 + 5 x 1 + 14 + 2 = 45 g / mol, since 4 molecules are 180 g / mol
3) Vm (O2) = 22.4 x 15 = 336 l / mol
4) 100 g: 180 g / mol = x L: 336 L / mol
x = (100 x 336): 180 = 186.7 l
Answer: V (O2) = 186.7 liters.