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

To solve the problem of how much calcium carbonate (CaCO₃) is needed to react completely with 25 mL of 0.75 M hydrochloric acid (HCl), we will follow these steps:

### Step 1: Calculate the number of moles of HCl

First, we need to calculate the number of moles of HCl present in the solution. The formula for calculating moles from molarity and volume is:

\[
\text{Moles of HCl} = \text{Molarity} \times \text{Volume (in L)}
...