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 )

Step-by-step solution

According to the above chemical reaction, 2 moles of bromine will give one mole of the product. 2 moles of Br 2   =   2 × 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   = 360 2 × 160   × 1 g   =   9 8   g   of   product But given that % yield = 27%. So actual amount