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» <html><body><p>`1.916g`<br/>`5.827g`<br/>`7.53g`<br/>`3.81g`</p>Solution :No. of Meq of HCl = `525xx(1)/(<a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a>)=52.5` <br/> No. of Meq of HCl = No. of mecq of `CaCl_(2)` = <br/> No. of Meq of <a href="https://interviewquestions.tuteehub.com/tag/plaster-601297" style="font-weight:bold;" target="_blank" title="Click to know more about PLASTER">PLASTER</a> of <a href="https://interviewquestions.tuteehub.com/tag/paris-11027" style="font-weight:bold;" target="_blank" title="Click to know more about PARIS">PARIS</a> <br/> So No. of Meq of plaster of paris = 52.5 <br/> `52.5=("wt of plaster of paris")/("GEW of plaste of paris")xx1000` <br/> `52.5=(<a href="https://interviewquestions.tuteehub.com/tag/x-746616" style="font-weight:bold;" target="_blank" title="Click to know more about X">X</a>)/(72.5)xx1000impliesx=3.81g`</body></html>