In bromination of Propyne, with Bromine 1, 1, 2, 2...

According to the above chemical reaction, 2 moles of bromine will give one mole of the product.

2 moles of ${Br}_{2} = 2 \times 160gof{Br}_{2}givesonemoleofproduct = 360gofproduct$

So, mass of product formed from $1gmof{Br}_{2}willbe = \dfrac{360}{2 \times 160} \times 1g = \dfrac{9}{8}gofproduct$

But given that % yield = 27%.

So actual amount of product $= \dfrac{9}{8} \times \dfrac{27}{100} = 0 . 30gram$

$= 3 . 0 \times {10}^{- 1}g$