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sneha.bagri

If 7.2g of bromine reacts wid 3.1g of K , d no.of moles of K required to combine wid Br2 to form KBr is???(KBr contains 34% by wt. of K)???

plz give ur working............

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 ...

RATE IF U FIND USEFUL...

sneha.bagri

@ akshay

d ans is 0.095

even i got d same ans as u got

akshay.khare91

me getting 0.075 and 0.09 is the answer the small difference can be neglected
because we have taken many things approx..like weight of K and Br
also avoided the decimal in finding 34% etc..

sneha.bagri

but d options given r

a)0.079
b)0.085
c)0.095
d)1

akshay.khare91

Taking the weight of Br as 74 gives the ans..

sneha.bagri

but wt. of Br is 79.9 ie. 80

akshay.khare91

soorryy taking weight 80 gives the ans..

the final weight of K will come out to be 3.61

hence mole = 3.61/38 = 0.095

cheerss!!!!!!!

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