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

Step 1. To calculate mass of HCl in 25 mL of 0.75 M Hcl
1000 mL of 0.75 M HCl contains HCl = 0.75 moles `=0.75xx36.5g=24.375g`
`therefore" 25 mL of 0.75 HCl will contain HCl"=(24.375)/(1000)xx25g=0.6844g.`
Step 2. To calculate mass of `CaCO_(3)` reacting completely with 0.9125 g of HCl
`CaCO_(3)(s)+2HCl(aq) rarr CaCl_(2)(aq)+CO_(2)(g)+H_(2)O(l)`
2 moles of HCl, i.e., `2xx36.5 g=73g HCl` react completely with `CaCO_(3) = "1 mole = 100 g"`
`therefore " 0.6844 g HCl will react completely with "CaCO_(3)=(100)/(73)xx0.6844g=0.938g`