Calculate the weight of 60% `H_(2)SO_(4)` required decomposing 50 g of chalk (Calcium carbonate ) .

Answer

To calculate the weight of 60% H₂SO₄ required to decompose 50 g of chalk (calcium carbonate, CaCO₃), we will follow these steps:

### Step 1: Write the balanced chemical equation
The reaction between sulfuric acid (H₂SO₄) and calcium carbonate (CaCO₃) can be represented as follows:

\[ \text{CaCO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{CaSO}_4 + \text{H}_2\text{O} + \text{CO}_2 \]

This shows that 1 mole of calcium carbonate reacts with 1 mole of sulfuric acid.

### Step 2: Calculate the molar mass of CaCO₃
The molar mass of calcium carbonate (CaCO₃) is calculated as follows:
- Calcium (Ca): 40.08 g/mol
- Carbon (C): 12.01 g/mol
- Oxygen (O): 16.00 g/mol × 3 = 48.00 g/mol

So, the molar mass of CaCO₃ = 40.08 + 12.01 + 48.00 = 100.09 g/mol.

### Step 3: Calculate the number of moles of CaCO₃ in 50 g
To find the number of moles of calcium carbonate in 50 g, use the formula:

\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} \]

\[ \text{Number of moles of CaCO}_3 = \frac{50 \text{ g}}{100.09 \text{ g/mol}} \approx 0.4995 \text{ moles} \]

### Step 4: Determine the amount of H₂SO₄ needed
According to the balanced equation, 1 mole of CaCO₃ requires 1 mole of H₂SO₄. Therefore, the number of moles of H₂SO₄ required is also approximately 0.4995 moles.

### Step 5: Calculate the mass of H₂SO₄ required
The molar mass of H₂SO₄ is:
- Hydrogen (H): 1.01 g/mol × 2 = 2.02 g/mol
- Sulfur (S): 32.07 g/mol
- Oxygen (O): 16.00 g/mol × 4 = 64.00 g/mol

So, the molar mass of H₂SO₄ = 2.02 + 32.07 + 64.00 = 98.09 g/mol.

Now, calculate the mass of H₂SO₄ needed:

\[ \text{Mass of H}_2\text{SO}_4 = \text{Number of moles} \times \text{Molar mass} \]

\[ \text{Mass of H}_2\text{SO}_4 = 0.4995 \text{ moles} \times 98.09 \text{ g/mol} \approx 49.0 \text{ g} \]

### Step 6: Adjust for the concentration of H₂SO₄
Since we have a 60% solution of H₂SO₄, we need to find out how much of this solution is required to obtain 49 g of pure H₂SO₄.

Let \( x \) be the mass of the 60% H₂SO₄ solution required:

\[ 0.60x = 49 \text{ g} \]

Solving for \( x \):

\[ x = \frac{49 \text{ g}}{0.60} \approx 81.67 \text{ g} \]

### Final Answer
The weight of 60% H₂SO₄ required to decompose 50 g of chalk is approximately **81.67 g**.

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