Calculate how much oxygen is required for the complete combustion of 30 g of methylamine?

Let’s implement the solution:
1. Let’s compose the reaction equation:
4СН3NH2 + 9O2 = 2N2 + 4CO2 + 10H2O – reaction of methylamine combustion, nitrogen and carbon dioxide are released;
2. Determine the number of moles of methylamine, if the mass is known: m (CH3NH2) = 30 g;
Y (CH3NH2) = m (CH3NH2) / M (CH3NH2); Y (CH3NH2) = 30/31 = 0.96 mol;
3. Let’s compose the proportion according to the reaction equation:
0.96 mol (CH3NH2) – X mol (O2);
-4 mol – 9 mol from here, X mol (O2) = 0.96 * 9/4 = 2.16 mol;
4. Determine the volume of oxygen, applying Avogadro’s law:
V (O2) = 2.16 * 22.4 = 48.38 liters.
Answer: for the combustion reaction of methylamine, oxygen is required in a volume of 48.38 liters.