How much marble of 96.5% purity would be required to prepare 100 litres of carbon dioxide at S.T.P. when marble is acted upon by dil HCl ?

Explanation

To determine how much marble of 96.5% purity is required to prepare 100 liters of carbon dioxide (CO2) at standard temperature and pressure (STP) when acted upon by dilute hydrochloric acid (HCl), we can follow these steps:

### Step 1: Determine the amount of CO2 produced
At STP, 1 mole of any gas occupies 22.4 liters. Therefore, to find out how many moles of CO2 are in 100 liters, we can use the formula:

\[ \text{Number of moles of CO2} = \frac{\text{Volume of CO2 at STP}}{22.4 \, \text{L/mol}} = \frac{100 \, \text{L}}{22.4 \, \text{L/mol}} \approx 4.464 \, \text{moles} \]

### Step 2: Relate moles of CO2 to moles of CaCO3
The reaction between marble (calcium carbonate, CaCO3) and dilute HCl produces CO2 as follows:

\[ \text{CaCO3} + 2 \text{HCl} \rightarrow \text{CaCl2} + \text{H2O} + \text{CO2} \]

From the balanced equation, we see that 1 mole of CaCO3 produces 1 mole of CO2. Thus, the number of moles of CaCO3 required is the same as the number of moles of CO2 produced:

\[ \text{Moles of CaCO3 required} = 4.464 \, \text{moles} \]

### Step 3: Calculate the mass of CaCO3 required
The molar mass of CaCO3 can be calculated as follows:
- Calcium (Ca) = 40 g/mol
- Carbon (C) = 12 g/mol
- Oxygen (O) = 16 g/mol (3 atoms)

\[ \text{Molar mass of CaCO3} = 40 + 12 + (16 \times 3) = 100 \, \text{g/mol} \]

Now, we can find the mass of CaCO3 required:

\[ \text{Mass of CaCO3} = \text{Moles} \times \text{Molar mass} = 4.464 \, \text{moles} \times 100 \, \text{g/mol} = 446.4 \, \text{grams} \]

### Step 4: Calculate the mass of marble required
Since the marble has a purity of 96.5%, we need to find the mass of the marble (x grams) that contains 446.4 grams of pure CaCO3:

\[ x \times \frac{96.5}{100} = 446.4 \, \text{grams} \]

To find x, we rearrange the equation:

\[ x = \frac{446.4 \times 100}{96.5} \approx 462.6 \, \text{grams} \]

### Conclusion
The mass of marble of 96.5% purity required to prepare 100 liters of carbon dioxide at STP is approximately **462.6 grams**.

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