Calculate the mass of 3% bromine water required to obtain 80 g of dibromoethane.

C2H6 + Br2 = C2H4Br2 + H2

Calculate the chemical amount of di-bromoethane: n = m / M

n (C2H4Br2) = 80 / (80 * 2 + 4 * 1 + 2 * 12) = 80/188 = 0.43 mol.

According to the reaction equation, since the ratio of the coefficients is 1: 1

n (Br2) = 0.43 mol.

Calculate the mass of pure bromine: m = n * M

m (Br2) = 0.43 * (80 * 2) = 0.43 * 160 = 68.8 gr.

We calculate the mass of bromine water: w = m (in-va) / m (pa-pa), therefore m (pa-pa) = m (in-va) / w, where w is substituted Not as a percentage, that is, w = 3 % / 100% = 0.03.

m (pa-pa) (Br2) = 68.8 / 0.03 = 229.33 gr.