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)
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Solution
75.20 In E2 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 ×0.2 mol= 0.4 mol Hence, the mass of the precipitate formed = Molar mass of AgBr× no. of moles formed = 188 × 0.4 g =75.20 g