Question

Calculate the maximum mass (in g) of $AgBr$ precipitate that will be formed when 0.2 mol of the starting material is taken in the above reaction? (Molar mass of $AgBr$=188 g/mol)

Answer 1

75.20
In $E_2$ mechanism, Anti elimination takes place. So, according to stereochemistry only 2 mol $KBr$ can be formed from 1 mol of the starting material. Hence,
By unitary method,
1 mol of starting material will give 2 mol of $KBr$
0.2 mol of starting material will give = 2 $\times 0.2$ mol= 0.4 mol
Hence, the mass of the precipitate formed = Molar mass of $AgBr$ $\times$ no. of moles formed = 188 $\times$ 0.4 g
=75.20 g