Calcium carbonate reacts with aqueous `HCl` to give `CaCl_(2)` and `CO_(2)` according to the reaction:
`CaCO_(3)(s)+2HCl(aq) rarr CaCl_(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

Let us calculate mass of HCl in 25 mL of 0.75 M HC1 1000 mL of 0.75 M HCl contain HCl = `0.75 xx 36.5` g 25 mL of 0.75 M HCl will contain HCl
`=(0.75 xx 36.5)/1000 xx 25 = 0.684`g
`underset(100 g)(CaCO_(3)) + underset(2 xx 36.5 g) (2HCl) to CaCl_(2) + CO_(2) + H_(2)O`
`2 xx 36.5`g of HCl react completely with `CaCO_(3)= 100g` 0.684 g of HCl will react completely with `CaCO_(3) = 100/(2 xx 36.5) xx 0.684 = 0.937`g