How much bromine can add to 30 g of butadiene-1.3?

m is the mass of the substance
v – amount of substance
Mr – molecular weight

Butadiene can attach a maximum of 2 bromine molecules, one bromine per double bond, and 1,2,3,4-tetrabromobutane is formed.

Let’s find the amount of butadiene substance by the formula v = m / Mr.
Mr (butadiene) = 54 g / mol
v (butadiene) = 30/54 = 0.56 mol.

Bromine is needed 2 times more, i.e. v (Br2) = 0.56 * 2 = 1.12 mol.

The mass is found by the formula m = Mr * v.
Mr (Br2) = 160 g / mol
m (Br2) = 1.12 * 160 = 179.2 g

Answer: 179.2 g.

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