How much marble of `96.5%` purity would be required to prepare 10 litres of carbon dioxide at STP when the marble is acted upon by dilute hydrochloric acid?

Explanation

To find out how much marble (calcium carbonate, CaCO₃) of 96.5% purity is required to prepare 10 liters of carbon dioxide (CO₂) at standard temperature and pressure (STP), we can follow these steps:

### Step 1: Understand the Reaction
The reaction between marble (CaCO₃) and dilute hydrochloric acid (HCl) can be represented as:
\[ \text{CaCO}_3 (s) + 2 \text{HCl} (aq) \rightarrow \text{CaCl}_2 (aq) + \text{CO}_2 (g) + \text{H}_2\text{O} (l) \]

From this reaction, we see that 1 mole of CaCO₃ produces 1 mole of CO₂.

### Step 2: Calculate Moles of CO₂ Required
At STP, 1 mole of any gas occupies 22.4 liters. Therefore, to find the number of moles of CO₂ in 10 liters:
\[ \text{Moles of CO}_2 = \frac{\text{Volume of CO}_2}{\text{Molar Volume at STP}} = \frac{10 \, \text{liters}}{22.4 \, \text{liters/mole}} \approx 0.4464 \, \text{moles} \]

### Step 3: Calculate Moles of CaCO₃ Required
Since 1 mole of CaCO₃ produces 1 mole of CO₂, the moles of CaCO₃ required will also be:
\[ \text{Moles of CaCO}_3 = 0.4464 \, \text{moles} \]

### Step 4: Calculate Mass of Pure CaCO₃ Required
The molar mass of CaCO₃ is approximately 100 g/mol. Therefore, the mass of pure CaCO₃ required is:
\[ \text{Mass of CaCO}_3 = \text{Moles} \times \text{Molar Mass} = 0.4464 \, \text{moles} \times 100 \, \text{g/mol} = 44.64 \, \text{grams} \]

### Step 5: Adjust for Purity
Since the marble is only 96.5% pure, we need to find the total mass of marble required to obtain 44.64 grams of pure CaCO₃:
Let \( X \) be the mass of marble needed.
\[ \frac{X \times 96.5}{100} = 44.64 \]
To find \( X \):
\[ X = \frac{44.64 \times 100}{96.5} \approx 46.26 \, \text{grams} \]

### Conclusion
Thus, the amount of marble of 96.5% purity required to prepare 10 liters of carbon dioxide at STP is approximately **46.26 grams**.

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