The mass of bromine interacting with ethylene 140 according to the reaction equation.

According to the condition of the problem, we write the equation:
Н2С = СН2 + Br2 = CH2Br – CH2Br – addition reaction, dibromoethane was obtained;
1 mol 1 mol;
Let’s make calculations using the formulas:
M (C2H4) = 28 g / mol;
M (Br2) = 159.8 g / mol;
Determine the amount of moles of ethylene by the formula:
Y (C2H4) = m / M = 140/28 = 5 mol;
In the equation, the amount of moles of ethylene and bromine is 1 mole, which means that Y (Br2) = 5 moles;
Let’s calculate the mass:
m (Br2) = Y * M = 5 * 159.8 = 799 g.
Answer: bromine is required for the reaction, its mass is exactly 799 g.