How much carbon dioxide and nitrogen will be released during the complete combustion of 6 liters of methylamine.

| September 23, 2021 | education

Find the amount of methylamine substance.

n = V: Vn.

n = 6 L: 22.4 L / mol = 0.27 mol.

Let’s compose the reaction equation, find the quantitative ratios of substances.

4CH3NH2 + 9O2 = 4CO2 + 2N2 + 10H2O.

According to the reaction equation, 4 mol of CH3NH2 accounts for 4 mol of CO2. Substances are in quantitative ratios 1: 1.

The amount of substance will be the same.

n (CO2) = n (CH3NH2) = 0.27 mol.

Find the volume of CO2.

V = Vn n, where Vn is the molar volume of gas, equal to 22.4 l / mol, and n is the amount of substance.

V = 0.2 7 mol × 22.4 L / mol = 6.048 L.

For 4 mol of CH3NH2, there are 2 mol of N2. The substances are in quantitative ratios of 1: 2.

The amount of substance N2 is 2 times less than CH3NH2.

n (N2) = ½ n (CH3NH2) = 0.27: 2 = 0.135 mol.

V = 0.135 mol × 22.4 L / mol = 3.024 L.

Answer: 3.024 L; 6.048 l.



One of the components of a person's success in our time is receiving modern high-quality education, mastering the knowledge, skills and abilities necessary for life in society. A person today needs to study almost all his life, mastering everything new and new, acquiring the necessary professional qualities.