# Calculate the mass of trichloromethane that can be obtained by reacting 5.6

**Calculate the mass of trichloromethane that can be obtained by reacting 5.6 liters of CH4, 28 liters of chlorine. Product yield 70% of theoretical.**

1. To solve, we compose the equations of the process:

V = 5.6 l; V = 28 l;

CH4 + Cl2 = CH3Cl + HCL – substitution occurs in the light;

CH3Cl + Cl2 = CH2Cl2 + HCl;

X g -? W = 70%

CH2Cl2 + Cl2 = CHCl3 + HCl – trichloromethane obtained.

2. Proportions:

1 mol of gas at normal level – 22.4 liters;

X mol (CH4) – 5.6 liters. hence, X mol (CH4) = 1 * 5.6 / 22.4 = 0.25 mol (the substance is in short supply).

Y (CH3Cl) = 0.25 mol since the amount of substances according to the equation is 1 mol.

1 mol of gas at normal level – 22.4 liters.

X mol (Cl2) – 28 L. hence, X mol (Cl2) = 1 * 28 / 22.4 = 1.25 mol (substance in excess).

Calculations are made for the substance in deficiency.

3. Find the mass of the product:

m (CH3Cl) = Y * M = 0.25 * 119.5 = 29.88 g (theoretical weight).

m (CH3Cl) = 0.70 * 29.88 = 20.9 g (practical weight).

Answer: Trichloromethane was obtained with a mass of 20.9 g.