The mass of 70% `H_(2)SO_(4)` required for neutralization of one mole of NaOH is:

Options

A. 49 gm
B. 98 gm
C. 70 gm
D. 34.3 gm

Answer

To find the mass of 70% H₂SO₄ required for the neutralization of one mole of NaOH, we can follow these steps:

### Step 1: Write the balanced chemical equation
The balanced chemical equation for the reaction between sulfuric acid (H₂SO₄) and sodium hydroxide (NaOH) is:
\[ \text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O} \]

### Step 2: Determine the mole ratio
From the balanced equation, we see that:
- 2 moles of NaOH react with 1 mole of H₂SO₄.

Thus, for 1 mole of NaOH, the amount of H₂SO₄ required is:
\[ \frac{1}{2} \text{ mole of H}_2\text{SO}_4 \]

### Step 3: Calculate the mass of H₂SO₄ required
The molar mass of H₂SO₄ is 98 g/mol. Therefore, the mass of 0.5 moles of H₂SO₄ is:
\[ \text{Mass of } \frac{1}{2} \text{ mole of H}_2\text{SO}_4 = \frac{98 \text{ g}}{2} = 49 \text{ g} \]

### Step 4: Calculate the mass of 70% H₂SO₄ solution required
Since we need to find the mass of a 70% H₂SO₄ solution that contains 49 g of pure H₂SO₄, we can set up the following equation:
\[ \text{Mass of 70% H}_2\text{SO}_4 = \frac{\text{Mass of pure H}_2\text{SO}_4}{\text{Percentage}} \times 100 \]
\[ \text{Mass of 70% H}_2\text{SO}_4 = \frac{49 \text{ g}}{70} \times 100 = 70 \text{ g} \]

### Final Answer
The mass of 70% H₂SO₄ required for the neutralization of one mole of NaOH is **70 g**.

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