Calcium carbonate reacts with aqueous HCl to give `CaCl_(2)` and `CO_(2)` according to the reaction, `CaCO_(3) (s) + 2 HCl (aq) to CaCl_(2) (aq) + CO_(2)(g) + H2O(l) `
What mass of CaCO3 is required to react completely with 25 mL of 0.75 M HCl?

Explanation

i. Mol of `HCl=(1000xx0.75)/1000=0.75`
Mass of `HCl=0.75xx36.5 g=24.375 g`
Wt. of `25 mL HCl=24.375/1000xx25=0.6844`
ii. According to equation:
`2 mol` pf `HCl` (i.e., `2xx36.5 g=73 g`)
`HCl` react completely with `CaCO_(3)=1 mol =100 g`.
`:. 0.6844 g HCl` will react completely with `CaCO_(3)`
`=100/73xx0.6844 g=0.938 g`