In an experiment,` 4.5` mol of calcium carbonate are reacted with dilute hydrochloric acid.
How many moles of HCI are used in this reaction?

Explanation

To solve the problem of how many moles of hydrochloric acid (HCl) are used when 4.5 moles of calcium carbonate (CaCO3) react with it, we will follow these steps:

### Step 1: Write the balanced chemical equation for the reaction.
The reaction between calcium carbonate and hydrochloric acid can be written as:

\[ \text{CaCO}_3 (s) + 2 \text{HCl} (aq) \rightarrow \text{CaCl}_2 (aq) + \text{CO}_2 (g) + \text{H}_2\text{O} (l) \]

### Step 2: Identify the mole ratio from the balanced equation.
From the balanced equation, we can see that:
- 1 mole of calcium carbonate (CaCO3) reacts with 2 moles of hydrochloric acid (HCl).

### Step 3: Calculate the moles of HCl required for 4.5 moles of CaCO3.
Since 1 mole of CaCO3 requires 2 moles of HCl, we can set up the calculation as follows:

\[ \text{Moles of HCl} = 2 \times \text{Moles of CaCO}_3 \]

Substituting the given value:

\[ \text{Moles of HCl} = 2 \times 4.5 = 9 \text{ moles of HCl} \]

### Final Answer:
Therefore, 9 moles of hydrochloric acid are used in the reaction with 4.5 moles of calcium carbonate.

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