What mass of bromine will take part in the reaction if 500 liters of ethylene are passed through bromine water.

What mass of bromine will take part in the reaction if 500 liters of ethylene are passed through bromine water. What is the mass and the number of moles of the reaction product.

Based on the data of the problem, we write down the equation of the process:
C2H4 + Br2 = C2H4Br2 – addition, dibromoethane is released;

We make calculations using the formulas:
M (C2H4) = 28 g / mol; M (Br2) = 159.8 g / mol; M (C2H4Br2) = 187.8 g / mol.

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

X mol (C2H2) – 500 liters from here, X mol (C2H2) = 1 * 500 / 22.4 = 22.3 mol;

Y (Br2) = 22.3 mol; Y (C2H4Br2) = 22.3 mol since the amount of substances is 1 mol.

We find the mass of bromine and product:
m (Br2) = Y * M = 22.3 * 159.8 = 3563.5 g = 3 kg 563 g

m (C2H4Br2) = Y * M = 22.3 * 187.8 = 4187.9 g = 4 kg 188 g

Answer: dibromoethane was obtained with a mass of 4 kg 188 g, in an amount of 22.3 mol, bromine with a mass of 3 kg 563 g is required