Question

A small amount of $CaC{O}_3$ completely neutralizes $525$ mL of $\frac{N}{10}HCl$ and no acid is left at the end. After converting all calcium chloride to $CaS{O}_4$, the amount of plaster of paris (in g) which can be obtained is (as nearest integer) :

Answer 1

Meq. of $CaC{O}_3=$ Meq. of $HCl$ $=$ Meq. of $Ca{Cl}_2$ formed
$\because$ Meq. of $HCl$ =$525\times \left(\frac{1}{10}\right)$ $=52.5$
$\therefore$ Meq. of $Ca{Cl}_2$ = $52.5$ = Meq. of $CaS{O}_4$ obtained by this $Ca{Cl}_2$
Also, Meq. of plaster of Paris ($CaS{O}_4\cdot \frac{1}{2}{H}_{2}O$) $=$Meq. of $CaS{O}_4$ obtained
or, $\frac{w}{145/2}\times 1000=52.5$
$\therefore$ Mass of $CaS{O}_4\cdot \frac{1}{2}{H}_{2}O=3.81$ g
So, the answer is $4$.