A small amount of CaCO_(3) completely neutralises 525 ml of (N)/(10) HCl and no acid is left at the end after converting all calcium chloride to CaSO_(4). How much plaster paris (CaSO_(4)(1)/(2)H_(2)O) can be obtained

A small amount of CaCO_(3) completely neutralises 525 ml of (N)/(10) H

1.	A small amount of CaCO_(3) completely neutralises 525 ml of (N)/(10) HCl and no acid is left at the end after converting all calcium chloride to CaSO_(4). How much plaster paris (CaSO_(4)(1)/(2)H_(2)O) can be obtained
Answer» `1.916g` `5.827g` `7.53g` `3.81g` Solution :No. of Meq of HCl = `525xx(1)/(10)=52.5` No. of Meq of HCl = No. of mecq of `CaCl_(2)` = No. of Meq of PLASTER of PARIS So No. of Meq of plaster of paris = 52.5 `52.5=("wt of plaster of paris")/("GEW of plaste of paris")xx1000` `52.5=(X)/(72.5)xx1000impliesx=3.81g`