How much bromine can add to 44.8 liters of ethene?

Let’s implement the solution:
Н2С = СН2 + Br2 = Br – CH2 – CH2 – Br – addition reaction, bromination of ethylene, dibromoethane was obtained;
Determine the molar mass of bromine:
M (Br2) = 2 * 79.9 = 159.8 g / mol;
Let’s calculate the number of moles of ethylene:
1 mol of gas at n. y – 22.4 l;
X mol (C2H4) – 44.8 L from here, X mol (C2H4) = 1 * 44.8 / 22.4 = 2 mol;
Let’s make a proportion according to the reaction equation:
2 mol (C2H4) – X mol (Br2);
-1 mol -1 mol hence, X mol (Br2) = 2 * 1/1 = 2 mol;
Let’s calculate the mass of bromine:
m (Br2) = Y * M = 2 * 159.8 = 319.6 g.
Answer: bromine weighing 319.6 g is needed to carry out the reaction.