What weight of `98%` pure `H_(2)SO_(4)` is required to neutralize `80gm` of `NaOH`?

Options

A. `98` gm
B. `0.98` gm
C. `100` gm
D. `980` gm

Explanation

To find the weight of 98% pure H₂SO₄ required to neutralize 80 g 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:

\[ \text{H}_2\text{SO}_4 + 2 \text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + 2 \text{H}_2\text{O} \]

This equation tells us that 1 mole of H₂SO₄ reacts with 2 moles of NaOH.

### Step 2: Calculate moles of NaOH
To find the number of moles of NaOH in 80 g, we need to use the molar mass of NaOH, which is 40 g/mol.

\[ \text{Moles of NaOH} = \frac{\text{mass}}{\text{molar mass}} = \frac{80 \text{ g}}{40 \text{ g/mol}} = 2 \text{ moles} \]

### Step 3: Determine moles of H₂SO₄ required
From the balanced equation, we know that 1 mole of H₂SO₄ is required for every 2 moles of NaOH. Therefore, for 2 moles of NaOH, we need:

\[ \text{Moles of H}_2\text{SO}_4 = \frac{2 \text{ moles NaOH}}{2} = 1 \text{ mole} \]

### Step 4: Calculate the mass of pure H₂SO₄ required
The molar mass of H₂SO₄ is 98 g/mol. Therefore, the mass of 1 mole of H₂SO₄ is:

\[ \text{Mass of H}_2\text{SO}_4 = 1 \text{ mole} \times 98 \text{ g/mol} = 98 \text{ g} \]

### Step 5: Adjust for purity
Since the H₂SO₄ is only 98% pure, we need to find out how much of the 98% solution is required to get 98 g of pure H₂SO₄. Let \( x \) be the mass of the 98% H₂SO₄ solution required.

\[ 0.98x = 98 \text{ g} \]

Solving for \( x \):

\[ x = \frac{98 \text{ g}}{0.98} = 100 \text{ g} \]

### Conclusion
The weight of 98% pure H₂SO₄ required to neutralize 80 g of NaOH is **100 g**.

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