Calcium Carbonate reacts with aqueous HCl to give `CaCl_(2)andCO_(2)` according to the reaction,
`CaCO_(3)(s)+2HCl(aq)toCaCl_(2)(aq)+CO_(2)(g)+H_(2)O(l)`
What mass of `CaCO_(3)` is required to react completely with 25 ml of 0.75 M HCl?

Explanation

`CaCO_(3)(s)+2HCl(aq)toCaCl_(2)(aq)+CO_(2)(g)+H_(2)O(l)`
Moles of HCl = `(25xx0.75)/1000=0.01875`
With 2 mol. Of HCl the mole of `CaCO_(3)` react is 1 mol. With 0.01875 mol. Of HCl the mole of `CaCO_(3)` that react is
`(0.01875xx1)/2=0.009375`
Wt. of `CaCO_(3)=0.009375xx100=0.9375`gm.