The amount of chlorine required for the complete chlorination of 40 liters of methane to carbon tetrachloride is …

Let’s implement the solution:

According to the condition of the problem, we compose the equations of the process:
CH4 + Cl2 = CH3Cl + HCl – substitution, occurs in the light, chloromethane is obtained;

CH3Cl + Cl2 = CH2Cl2 + HCl – dichloromethane formed;

CH2Cl2 + Cl2 = CHCl3 + HCl – trichloromethane is released;

CHCl3 + Cl2 = CCl4 + HCl – the final product is carbon tetrachloride.

Proportion:
1 mol of gas at normal level – 22.4 l

X mol (CH4) -40 L from here, X mol (CH4) = 1 * 40 / 22.4 = 1.8 mol.

Y (CCl4) = 1.8 mol since the amount of substances is 1 mol.

Find the volume of the product:
V (CCl4) = 1.8 * 22.4 = 40.32 L

Answer: the volume of carbon tetrachloride is 40.32 liters