Calcium carbonate reacts with aqueous HCl to give `CaCl_(2)` and `CO_(2)` according to the reaction given below
`CaCO_(3)(s)+2HCl(aq)rarrCaCl_(2)(aq)+CO_(2)(g)+H_(2)O(l)`
What mass of `CaCl_(2)` will be formed when 250mL of 0.76 M HCl reacts with 1000 g of `CaCO_(3)`? Name the limiting reagent. Calculate the number of moles of `CaCl_(2)` formed in the reaction.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Write the balanced chemical equation
The balanced equation for the reaction of calcium carbonate (CaCO₃) with hydrochloric acid (HCl) is:
\[ \text{CaCO}_3(s) + 2 \text{HCl}(aq) \rightarrow \text{CaCl}_2(aq) + \text{CO}_2(g) + \text{H}_2O(l) \]

### Step 2: Calculate the moles of CaCO₃
To find the number of moles of CaCO₃, we use the formula:
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