Chemical Engineering - Stoichiometry - Discussion

Firstly 80 kg of Na2So4 are present in an aqueous solution & the total weight of the solution is 330 kg.
After cooling Na2So4.10H2O crystals are separated whose total weight is 80 kg.

Now,
Molecular weight of
Na2So4.10H2O = 142 + 10(18) = 322

Now calculate the amount of Na2So4 in separated crystals.
322 kg Na2So4.10H2O = 142 kg Na2So4
80 kg Na2So4.10H2O = ?

So, we get 35.28 kg of Na2So4 which is present in separated crystals.

Now originally 80 kg of Na2So4 are present.
So, Na2So4 available in the remaining solution;
= 80 - 35.28 = 44.72 kg.

Now finally the weight fraction of Na2So4 in the remaining solution
=weight of Na2So4 in sol / Total weight of sol.
= 44.72/[ 330 - 80 ]
= 0.179 approx. 0.18.

Percentage of Na2SO4 in Na2So4.10 H2O = 142/(142+180).
So in 80 kg it is 80*(142/(322) = 35.
Therefore Na2SO4 is 80-35 = 44.
So percentage remaining is 44/330-80) = 0.18.