`62.5` gm of a mixture of `CaCO_(3)` and `SiO_(2)` are treated with excess of `HCl` and `1.1` gm of `CO_(2)` is produced. What is mass `%` `CaCO_(3)` in the mixture.

Explanation

To find the mass percentage of \( \text{CaCO}_3 \) in the mixture, we can follow these steps:

### Step 1: Understand the Reaction
When \( \text{CaCO}_3 \) reacts with hydrochloric acid (\( \text{HCl} \)), it produces calcium chloride (\( \text{CaCl}_2 \)), carbon dioxide (\( \text{CO}_2 \)), and water (\( \text{H}_2\text{O} \)). The reaction can be represented as:
\[ \text{CaCO}_3 + 2 \text{HCl} \rightarrow \text{CaCl}_2 + \text{CO}_2 + \text{H}_2\text{O} \]
From this reaction, we can see that 1 mole of \( \text{CaCO}_3 \) produces 1 mole of \( \text{CO}_2 \).
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