Find the volume of chlorine required to interact with ethylene with a volume of 4.48 liters.

To solve the problem, we write the equation:
Н2С = СН2 + Cl2 = CH2Cl – CH2Cl – the reaction of addition of chlorine at the site of the breaking of the double bond in ethylene, dichloroethane is released;
Determine the amount of moles of ethylene:
1 mol of gas at normal level – 22.4 liters.
X mol (C2H4) – 4.48 L. hence, X mol (C2H4) = 1 * 4.48 / 22.4 = 0.2 mol;
According to the reaction equation, the amount of moles of ethylene and chlorine is equal to 1 mol, which means that Y (Cl2) = 0.2 mol.
Find the amount of chlorine:
V (Cl2) = 0.2 * 22.4 = 4.48 liters.
Answer: the volume of chlorine is 4.48 liters.