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

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.