# Calculate the mass of the reaction product, which is obtained by the interaction of ethylene with a volume

**Calculate the mass of the reaction product, which is obtained by the interaction of ethylene with a volume of 1.12 liters with a bromine mass of 1.6 g. Call this substance.**

To solve, write the process equation:

H2C = CH2 + BR2 = BR – CH2 – CH2 – BR (C2H4BR) – accession, Diberomethane was formed;

Calculations:

M (BR2) = 159.8 g / mol;

M (C2H4BR2) = 187.8 g / mol.

Proportion:

1 mol of gas at n. – 22.4 l;

X mol (C2H4) – 1.12 liters from here, x mol (C2H4) = 1 * 1,12 / 22.4 = 0.05 mol (substance in excess).

We define the amount of starting material:

Y (br2) = m / m = 1.6 / 159.8 = 0.01 mol (substance in disadvantage);

Y (C2H4Br2) = 0.01 mol as the number of these substances is 1 mol.

Calculations produce by substance in a shortage.

We find a lot of product:

M (C2H4BR2) = Y * m = 0.01 * 187.8 = 1.88 g

Answer: The mass of Diberomethane is 1.88 g