# 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. 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.