Cho 39,4 gam BaCO3 tác dụng với 150g dung dịch H2SO4, sau phản ứng thu được muối BaSO4, khí CO2 và H₂O

To solve this problem, we will analyze the reaction step by step to determine the products and their quantities.

### Given Reaction:
1. **BaCO₃ + H₂SO₄ → BaSO₄ + CO₂ + H₂O**

### 1. Calculate moles of BaCO₃ used:
- Molar mass of BaCO₃ = 137 (Ba) + 12 (C) + 3 × 16 (O) = 197 g/mol
- Moles of BaCO₃ = \( \frac{39.4 \text{ g}}{197 \text{ g/mol}} \approx 0.2 \text{ mol} \)

### 2. Reaction stoichiometry:
From the balanced equation, we see that 1 mole of BaCO₃ reacts with 1 mole of H₂SO₄ to produce 1 mole of BaSO₄, 1 mole of CO₂, and 1 mole of H₂O.

### 3. Calculate the amount of H₂SO₄ required:
Assuming the complete reaction of BaCO₃ with H₂SO₄:
- Moles of H₂SO₄ needed = Moles of BaCO₃ = 0.2 mol

### 4. Concentration of H₂SO₄ solution:
- Given: Weight of H₂SO₄ solution = 150 g
- Assuming a standard concentration of H₂SO₄ (e.g., 98 g/mol for pure H₂SO₄):
- Mass of H₂SO₄ in 150 g solution = \( \frac{98 \text{ g/mol}}{\text{density of solution}}\) (Density needs to be known for precise calculation; typically around 1.84 g/mL for concentrated solutions.)

Assuming \( \text{density} = 1.84 \, \text{g/mL} \):
- Volume of solution = \( \frac{150 \text{ g}}{1.84 \text{ g/mL}} \approx 81.52 \, \text{mL}\)
- Moles of H₂SO₄ = \( \frac{150 \text{ g}}{98 \text{ g/mol}} \approx 1.53 \, \text{mol} \)

### 5. Products:
- Products from the reaction are CO₂, H₂O, and BaSO₄.
- Since 0.2 moles of BaCO₃ react selectively, they will produce:
- 0.2 moles CO₂
- 0.2 moles H₂O
- 0.2 moles BaSO₄

### 6. Volume of CO₂ produced:
Using the ideal gas law at STP:
- Volume of 1 mole of gas at STP = 22.4 L
- Therefore, volume of CO₂ = \( 0.2 \, \text{mol} \times 22.4 \, \text{L/mol} \approx 4.48 \, \text{L} \)

### Final Summary:
- **a)** Write the PTHH (reaction equation).
- **b)** Volume of CO₂ produced is approximately 4.48 L.
- **c)** Calculate any remaining amounts, if necessary.

This solution should give you a clear understanding of the reaction and the calculations involved. Let me know if you need further assistance!