Answer :
Answer:
PLEASE MARK AS BRAINLIEST
Explanation:
To determine the volume of CO2 liberated when 20g of 50% pure calcium carbonate (\( \text{CaCO}_3 \)) is treated with excess dilute sulfuric acid (\( \text{H}_2\text{SO}_4 \)), we follow these steps:
1. **Calculate moles of \( \text{CaCO}_3 \)**:
Molecular weight of \( \text{CaCO}_3 \):
\( \text{Ca} = 40.08 \) g/mol,
\( \text{C} = 12.01 \) g/mol,
\( \text{O}_3 = 3 \times 16.00 = 48.00 \) g/mol.
Therefore, molecular weight of \( \text{CaCO}_3 \):
\( 40.08 + 12.01 + 48.00 = 100.09 \) g/mol.
Moles of \( \text{CaCO}_3 \) in 20g:
\( \text{Moles} = \frac{\text{Mass}}{\text{Molecular weight}} = \frac{20 \text{ g}}{100.09 \text{ g/mol}} = 0.1999 \text{ mol} \approx 0.2 \text{ mol} \).
2. **Reaction of \( \text{CaCO}_3 \) with \( \text{H}_2\text{SO}_4 \)**:
\( \text{CaCO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{CaSO}_4 + \text{CO}_2 + \text{H}_2\text{O} \)
According to the reaction, 1 mole of \( \text{CaCO}_3 \) produces 1 mole of \( \text{CO}_2 \).
3. **Calculate moles of \( \text{CO}_2 \)**:
Moles of \( \text{CO}_2 \) liberated = 0.2 mol.
4. **Apply ideal gas law to find volume**:
Given conditions:
- Pressure (\( P \)) = 1 atm
- Temperature (\( T \)) = 273 K
- Volume (\( V \)) = ?
Ideal gas law: \( PV = nRT \)
Where:
\( R \) is the universal gas constant (0.0821 L·atm/mol·K).
\( V = \frac{nRT}{P} \)
Substitute the values:
\( V = \frac{0.2 \text{ mol} \times 0.0821 \text{ L·atm/mol·K} \times 273 \text{ K}}{1 \text{ atm}} \)
\( V = \frac{4.47826 \text{ L·atm}}{1} \)
\( V \approx 4.48 \text{ L} \)
5. **Conclusion**:
The volume of \( \text{CO}_2 \) liberated at 1 atm pressure and 273 K when 20g of 50% pure \( \text{CaCO}_3 \) is treated with excess dilute \( \text{H}_2\text{SO}_4 \) is approximately \( \boxed{4.48 \text{ L}} \).
Answer:
Baldeo's duty was to check if the tunnel is obstruction-free before the arrival of the mail train. He also checked if the signal lamp is burning at night. His son asked, "Shall I come too, father?"