A small amount of CaCO3 completely neutralizes
525ml of .1N HCl and no acid is left in the end.
After converting all calcium chloride to CaSO4, how much plaster of paris can be obtained?

Question

A small amount of CaCO3 completely neutralizes 
525ml of .1N HCl and no acid is left in the end.
After converting all calcium chloride to CaSO4, how much plaster of paris can be obtained?

anonymous · Answer

First write a balanced equation.

anonymous · Answer

There is actually no need!!

anonymous · Answer

CaCO3 + HCl -> HCO3 + CaCl
CaCl + H2SO4 -> CaSO4 + H+ + Cl-
CaSO4 + H+ Cl- -> Ca2H2O9S2 + Cl-

That's my attempt to make sense of the problem..

anonymous · Answer

In step 2, to be balanced it would be 2H+ on the product side.

anonymous · Answer

In step 3, to be balanced it would be 2H+ on the reactant side.

anonymous · Answer

Correct me if I'm wrong with the mechanism.

anonymous · Answer

What i think....
525ml of .1N HCl = 525ml of 0.1N CaCl2 = 525ml of .1N of PoP

Equivalent mass of PoP = 145/2 g = 72.5g

THUS, Mass of PoP in 525 ml of .1N solution
= normality * Eq mass * Volume/1000
= .1*72.5*525/1000
= 3.806g!