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 . (Molar Mass : Bromine = 80   g / mol )

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 &#160; g of Bromine in this reaction is_____ &#215; 10 - 1 &#160; g . (Molar Mass : Bromine = 80 &#160; g / mol )

Aicharya · Accepted Answer

According to the above chemical reaction, 2 moles of bromine will give one mole of the product.
2 moles of ${Br}_{2} = 2 \times 160 g of {Br}_{2} gives one mole of product = 360 g of product$
So, mass of product formed from $1 gm of {Br}_{2} will be = \frac{360}{2 \times 160} \times 1 g = \frac{9}{8} g of product$
But given that % yield = 27%.
So actual amount of product $= \frac{9}{8} \times \frac{27}{100} = 0.30 g r a m$
$= 3.0 \times 10^{- 1} g$

In bromination of Propyne, with Bromine 1 , 1 , 2 , 2 -tetrabromopropane is obtained in 27 % yield. The…

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