# Chlorination of 56 liters of methane (volume normalized) yielded 239 g of chloroform

**Chlorination of 56 liters of methane (volume normalized) yielded 239 g of chloroform. Determine the mass fraction of the reaction product yield.**

1. To solve, we compose the equations:

CH4 + Cl2 = CH3Cl + HCl – substitutions;

CH3Cl + Cl2 = CH2Cl2 + HCl;

CH2Cl2 + Cl2 = CHCl3 + HCl – chloroform obtained;

2. Calculation:

M (CHCl3) = 119.5 g / mol.

3. Determine the amount of the original substance:

1 mol of gas at normal level – 22.4 liters;

X mol (CH4) – 56 liters. hence, X mol (CH4) = 1 * 56 / 22.4 = 2.5 mol;

Y (CHCl3) = 2.5 mol since the number of moles of these substances according to the equation is 1 mol.

4. Find the mass and yield of the product:

m (CHCl3) = Y * M = 2.5 * 119.5 = 298.75 g;

W = m (practical) / m (theoretical) * 100;

W = 239 / 298.75 * 100 = 80%

Answer: the chloroform yield was 80%.