PASSAGE-IV : 10 g of NaOH required certain amount of `H_(2)SO_(4)` for complete neutralisation.
Calculate the amount of `H_(2)SO_(4)`.

Options

A. 10 g
B. 12.24 g
C. 24.5 g
D. 49 g

Explanation

To solve the problem of how much \( H_2SO_4 \) is required to completely neutralize 10 g of \( NaOH \), we can follow these steps:

### Step 1: Write the balanced chemical equation for the neutralization reaction.
The balanced reaction between sodium hydroxide (\( NaOH \)) and sulfuric acid (\( H_2SO_4 \)) is:
\[ 2 \, NaOH + H_2SO_4 \rightarrow Na_2SO_4 + 2 \, H_2O \]

### Step 2: Determine the molar masses of \( NaOH \) and \( H_2SO_4 \).
- Molar mass of \( NaOH \):
- Sodium (Na): 23 g/mol
- Oxygen (O): 16 g/mol
- Hydrogen (H): 1 g/mol
- Total: \( 23 + 16 + 1 = 40 \, g/mol \)

- Molar mass of \( H_2SO_4 \):
- Hydrogen (H): 1 g/mol (2 H)
- Sulfur (S): 32 g/mol
- Oxygen (O): 16 g/mol (4 O)
- Total: \( 2 + 32 + (4 \times 16) = 2 + 32 + 64 = 98 \, g/mol \)

### Step 3: Use stoichiometry to find the amount of \( H_2SO_4 \) needed.
From the balanced equation, we see that:
- 2 moles of \( NaOH \) react with 1 mole of \( H_2SO_4 \).

### Step 4: Calculate the number of moles of \( NaOH \) in 10 g.
\[ \text{Moles of } NaOH = \frac{\text{mass}}{\text{molar mass}} = \frac{10 \, g}{40 \, g/mol} = 0.25 \, moles \]

### Step 5: Determine the moles of \( H_2SO_4 \) required for 0.25 moles of \( NaOH \).
From the stoichiometry:
- 2 moles of \( NaOH \) require 1 mole of \( H_2SO_4 \).
- Therefore, 0.25 moles of \( NaOH \) will require:
\[ \text{Moles of } H_2SO_4 = \frac{0.25 \, moles \, NaOH}{2} = 0.125 \, moles \, H_2SO_4 \]

### Step 6: Convert moles of \( H_2SO_4 \) to grams.
\[ \text{Mass of } H_2SO_4 = \text{moles} \times \text{molar mass} = 0.125 \, moles \times 98 \, g/mol = 12.25 \, g \]

### Final Answer:
The amount of \( H_2SO_4 \) required to completely neutralize 10 g of \( NaOH \) is **12.25 g**.

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