In an experiment,` 4.5` mol of calcium carbonate are reacted with dilute hydrochloric acid.
What mass of calcium chloride is formed? (Relative molecular mass of calcium chloride is 111.)

Explanation

To solve the problem of determining the mass of calcium chloride formed from the reaction of calcium carbonate with dilute hydrochloric acid, we will follow these steps:

### Step 1: Write the balanced chemical equation
The balanced chemical equation for the reaction between calcium carbonate (CaCO₃) and hydrochloric acid (HCl) is:

\[ \text{CaCO}_3 + 2 \text{HCl} \rightarrow \text{CaCl}_2 + \text{CO}_2 + \text{H}_2\text{O} \]

### Step 2: Identify the mole ratio
From the balanced equation, we can see that:
- 1 mole of CaCO₃ produces 1 mole of CaCl₂.

### Step 3: Calculate the moles of CaCl₂ produced
Given that we have 4.5 moles of calcium carbonate (CaCO₃), the moles of calcium chloride (CaCl₂) produced will also be 4.5 moles, since the mole ratio is 1:1.

\[ \text{Moles of CaCl}_2 = 4.5 \text{ moles} \]

### Step 4: Use the molar mass to find the mass of CaCl₂
The relative molecular mass of calcium chloride (CaCl₂) is given as 111 g/mol. To find the mass of calcium chloride produced, we use the formula:

\[ \text{Mass} = \text{Moles} \times \text{Molar Mass} \]

Substituting the values we have:

\[ \text{Mass of CaCl}_2 = 4.5 \text{ moles} \times 111 \text{ g/mol} = 499.5 \text{ g} \]

### Step 5: Conclusion
The mass of calcium chloride (CaCl₂) formed from the reaction is **499.5 grams**.

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