How many grams of dichloroethane can be obtained by the interaction of 4 liters of ethane and 3.4 liters of chlorine?

We carry out the solution according to the reaction equation:
C2H6 + Cl2 = C2H5Cl + HCl – substitution reaction, chloroethane was obtained;
M (C2H5Cl) = 64.5 g / mol.
Determine the amount of moles of ethane, chlorine:
1 mol of gas at n, y – 22.4 liters.
X mol (C2H6) – 4 liters. hence, X mol (C2H6) = 1 * 4 / 22.4 = 0.178 mol (substance in excess);
1 mol of gas at n, y – 22.4 liters.
X mol (Cl2) – 3.4 liters from here, X mol (Cl2) = 1 * 3.4 / 22.4 = 0.15 mol (the substance is in short supply);
Further calculations are made for the substance in deficiency.
According to the reaction equation, the amount of mol of chlorine and chloroethane is equal to 1 mol, which means that Y (C2H5Cl) = 0.15 mol.
Let’s calculate the mass of chloroethane:
m (C2H5Cl) = Y * M = 0.15 * 64.5 = 9.67 g.
Answer: the mass of chloroethane is 9.67 g.