Question

In bromination of Propyne, with Bromine 1,1,2,2-tetrabromopropane is obtained in 27% yield. The amount of 1,1,2,2-tetrabromopropane obtained from 1 g of Bromine in this reaction is _______ × 10^–1 g. (Nearest integer)
(Molar Mass : Bromine = 80 g/mol)

Answer 1

Correct Answer: 3

2 moles Br₂=1 mole 1,1,2,2-tetrabromopropane
\(\frac{1}{160} \, \text{mole Br}_2 = \frac{1}{2} \times \frac{1}{160} \, \text{mole 1,1,2,2-tetrabromopropane}\)
But yield of reaction is only 27%
Moles of 1,1,2,2-tetrabromopropane \(=\)\(\frac{1}{2} \times \frac{1}{160} \times \frac{27}{100}\)
Molar mass of 1,1,2,2-tetrabromopropane = 360 g
Mass of 1,1,2,2-tetrabromopropane\(=\)\(12 \times \frac{1}{160} \times \frac{27}{100} \times 360 \, \text{g}\)
\(= 3 × 10^{–1} \text{g}\)
So, the answer is 3.