How much air is needed to burn 8L of methylamine?

n = V: Vn

n = 8 L: 22.4 L / mol = 0.36 mol.

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

4СН3NН2 + 9O2 = 4СО2 ↑ + 10H2O + 2N2 ↑.

For 4 mol of methylamine, there are 9 mol of oxygen. The substances are in quantitative ratios 4: 9 = 1: 2.25. The amount of oxygen will be 2.5 times more than the amount of methylamine.

n (O2) = 2.5 n (CH3NH2) = 2.25 × 0.36 = 0.81 mol.

Let’s find the volume of oxygen.

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 = 22.4 L / mol × 0.81 mol = 18.144 L.

18.144 L – 21%,

Vair – 100%?

Vair = (18.144 × 100%): 21% = 86.4 l = 90 l.

Answer: V = 90 liters.



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