When methylamine burns, CO2 and H2O and 1.12 liters of N2 are formed. Determine

When methylamine burns, CO2 and H2O and 1.12 liters of N2 are formed. Determine: a) the mass of burned methylamine. b) volume of CO2 c) mass of water

1. Let us write down the equation according to the problem statement:
4CH3NH2 + 9O2 = 4CO2 + 10H2O + 2N2 – OBP, carbon dioxide, nitrogen, water are released;
2. Calculation by formulas:
M (CH3NH2) = 43 g / mol;
M (CO2) = 44 g / mol;
M (H2O) = 18 g / mol.
3. Let’s calculate the number of moles of nitrogen, carbon dioxide:
1 mole of gas at normal level – 22.4 liters;
X mol (N2) – 1.12 L. hence, X mol (N2) = 1 * 1.12 / 22.4 = 0.05 mol (deficient substance).
The calculation is made for the substance in deficiency.
X mol (CO2) – 0.05 mol (N2);
-4 mol – 2 mol from here, X mol (CO2) = 0.05 * 4/2 = 0.1 mol.
4. Determine the number of moles of methylamine, water:
X mol (CH3NH2) – 0.05 mol (N2);
-4 mol – 2 mol from here, X mol (CH3NH2) = 4 * 0.05 / 2 = 0.1 mol;
X mol (H2O) – 0.05 mol (N2);
-10 mol – 2 mol from here, X mol (H2O) = 10 * 0.05 / 2 = 0.25 mol.
5. Find the mass of water:
m (H2O) = 0.25 * 18 = 4.5 g.
carbon monoxide volume:
V (CO2) = 0.1 * 22.4 = 2.24 l.
Methylamine weight:
m (CH3NH2) = Y * M = 0.1 * 43 = 4.3 g.
Answer: to carry out the reaction, methylamine with a mass of 4.3 g is required, the products are obtained: V (CO2) = 2.24 l, m (H2O) = 4.5 g.