10 g of a piece of marble was put into excess of dilute HCl acid. When the reaction was complete, `1120cm^3` of `CO_2` was obtained at STP. The percentage of `CaCO_3` in the marble is

Options

A. `25%`
B. `50%`
C. `75%`
D. `100%`

Answer

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

### Step 1: Write the balanced chemical equation
The reaction between marble (calcium carbonate, \( \text{CaCO}_3 \)) and hydrochloric acid (\( \text{HCl} \)) can be represented as:
\[ \text{CaCO}_3 (s) + 2 \text{HCl} (aq) \rightarrow \text{CaCl}_2 (aq) + \text{H}_2O (l) + \text{CO}_2 (g) \]

### Step 2: Determine the volume of \( \text{CO}_2 \) produced
From the problem, we know that \( 1120 \, \text{cm}^3 \) of \( \text{CO}_2 \) was produced at STP.

### Step 3: Convert the volume of \( \text{CO}_2 \) to moles
At STP, 1 mole of gas occupies \( 22.4 \, \text{L} \) or \( 22400 \, \text{cm}^3 \). Therefore, we can calculate the number of moles of \( \text{CO}_2 \) produced:
\[ \text{Moles of } \text{CO}_2 = \frac{1120 \, \text{cm}^3}{22400 \, \text{cm}^3/\text{mol}} = \frac{1120}{22400} = 0.05 \, \text{mol} \]

### Step 4: Relate moles of \( \text{CO}_2 \) to moles of \( \text{CaCO}_3 \)
From the balanced equation, we see that 1 mole of \( \text{CaCO}_3 \) produces 1 mole of \( \text{CO}_2 \). Therefore, the moles of \( \text{CaCO}_3 \) that reacted is also \( 0.05 \, \text{mol} \).

### Step 5: Calculate the mass of \( \text{CaCO}_3 \)
The molar mass of \( \text{CaCO}_3 \) can be calculated as follows:
- Calcium (Ca) = 40 g/mol
- Carbon (C) = 12 g/mol
- Oxygen (O) = 16 g/mol × 3 = 48 g/mol

Thus, the molar mass of \( \text{CaCO}_3 \) is:
\[ 40 + 12 + 48 = 100 \, \text{g/mol} \]

Now, we can calculate the mass of \( \text{CaCO}_3 \) that reacted:
\[ \text{Mass of } \text{CaCO}_3 = \text{Moles} \times \text{Molar mass} = 0.05 \, \text{mol} \times 100 \, \text{g/mol} = 5 \, \text{g} \]

### Step 6: Calculate the percentage of \( \text{CaCO}_3 \) in the marble
The percentage of \( \text{CaCO}_3 \) in the marble is given by:
\[ \text{Percentage of } \text{CaCO}_3 = \left( \frac{\text{Mass of } \text{CaCO}_3}{\text{Total mass of marble}} \right) \times 100 = \left( \frac{5 \, \text{g}}{10 \, \text{g}} \right) \times 100 = 50\% \]

### Conclusion
The percentage of \( \text{CaCO}_3 \) in the marble is **50%**.

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