IIT JEE , AIEEE Forum - Chemistry , Physical Chemistry

akshay.khare91

from balanced chemical reaction we see that

140 f of Br2 reacts with 76g of K (taking Br = 70g and K= 38g)

therefore 7.2g of Br2 will react with = 3.9g of K

hence K is limiting reagent which will decide quantity of product

again form balanced chemical reaction

76g of K gives 216g of KBr

hence 3.1 g of K will give = 8.81g of KBr ..

now since weight of KBr = 108 (approx.. )
and it contains 34% of K = 36.72g of K

Since 108g of KBr contain 36.72 g of K
therefore 8.81 g of Kbr will contain = 2.99g of K

hence no. of mole = weight / molecular weight
= 2.99/38 = 0.079 mole of K will be required ...

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