Let’s write the reaction equation:
C6H6 + Br2 (cat.) = C6H5Br + HBr
Here iron (III) bromide, FeBr3 can be used as a catalyst (cat.).
Let’s find the amount of benzene substance:
v (C6H6) = m (C6H6) / M (C6H6) = 156/78 = 2 (mol).
According to the reaction equation, from 1 mol of benzene, 1 mol of bromobenzene is formed, therefore, at 100% yield of the reaction product:
v (C6H5Br) = v (C6H6) = 2 (mol).
Thus, the theoretical mass of the resulting bromobenzene:
m (C6H5Br) = v (C6H5Br) * M (C6H5Br) = 2 * 157 = 314 (g)
and its output:
Ф (C6H5Br) = (mpract. (C6H5Br) / m (C6H5Br)) * 100 = (160/314) * 100 = 50.95 = 51 (%).
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