Calculate the amount of 50% `H_(2)SO_(4)` required to decompose 25g calcium carbonate :-

Options

A. 98gm
B. 49gm
C. 24.5gm
D. 196gm

Explanation

To calculate the amount of 50% H₂SO₄ required to decompose 25g of 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:
\[ \text{H}_2\text{SO}_4 + \text{CaCO}_3 \rightarrow \text{CaSO}_4 + \text{H}_2\text{O} + \text{CO}_2 \]

### Step 2: Determine the molar masses
- Molar mass of H₂SO₄ = 2(1) + 32 + 4(16) = 98 g/mol
- Molar mass of CaCO₃ = 40 + 12 + 3(16) = 100 g/mol

### Step 3: Establish the mole ratio
From the balanced equation, we see that 1 mole of H₂SO₄ reacts with 1 mole of CaCO₃. Therefore, the molar ratio is:
\[ 1 \text{ mol H}_2\text{SO}_4 : 1 \text{ mol CaCO}_3 \]

### Step 4: Calculate the moles of CaCO₃
To find the moles of CaCO₃ in 25g:
\[ \text{Moles of CaCO}_3 = \frac{\text{mass}}{\text{molar mass}} = \frac{25 \text{ g}}{100 \text{ g/mol}} = 0.25 \text{ mol} \]

### Step 5: Calculate the amount of H₂SO₄ required
Since the mole ratio is 1:1, we need 0.25 moles of H₂SO₄ to react with 0.25 moles of CaCO₃. Now, we can calculate the mass of H₂SO₄ required:
\[ \text{Mass of H}_2\text{SO}_4 = \text{moles} \times \text{molar mass} = 0.25 \text{ mol} \times 98 \text{ g/mol} = 24.5 \text{ g} \]

### Step 6: Calculate the amount of 50% H₂SO₄ solution required
Since we are using a 50% H₂SO₄ solution, this means that in every 100g of the solution, there are 50g of H₂SO₄. To find out how much of the 50% solution is needed to get 24.5g of H₂SO₄, we can set up a proportion:
\[ \frac{50 \text{ g H}_2\text{SO}_4}{100 \text{ g solution}} = \frac{24.5 \text{ g H}_2\text{SO}_4}{x \text{ g solution}} \]

Cross-multiplying gives:
\[ 50x = 24.5 \times 100 \]
\[ 50x = 2450 \]
\[ x = \frac{2450}{50} = 49 \text{ g} \]

### Final Answer
Thus, the amount of 50% H₂SO₄ required to decompose 25g of calcium carbonate is **49 grams**.

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