Sulphuric acid reacts with sodium hydroxide as follows
`H_(2)SO_(4)+2NaOHrarrNa_(2)SO_(4)+2H_(2)O`
when 1L of 0.1M sulphuric acid solution is allowed to react with 1L of 0.1M sodium hydroxide solution, the amount of sodium solphate formed and its molarity in the solution obtained is

Options

A. `0.1"mol"L^(-1)`
B. `7.10g`
C. `0.025"mol"L^(-1)`
D. 3.55g

Explanation

To solve the problem, we need to follow these steps:

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

### Step 2: Calculate the moles of H₂SO₄ and NaOH
Given:
- Volume of H₂SO₄ solution = 1 L
- Molarity of H₂SO₄ = 0.1 M
- Volume of NaOH solution = 1 L
- Molarity of NaOH = 0.1 M

Using the formula:
\[ \text{Moles} = \text{Molarity} \times \text{Volume (L)} \]

For H₂SO₄:
\[ \text{Moles of H}_2\text{SO}_4 = 0.1 \, \text{mol/L} \times 1 \, \text{L} = 0.1 \, \text{moles} \]

For NaOH:
\[ \text{Moles of NaOH} = 0.1 \, \text{mol/L} \times 1 \, \text{L} = 0.1 \, \text{moles} \]

### Step 3: Determine the limiting reagent
From the balanced equation, 1 mole of H₂SO₄ reacts with 2 moles of NaOH. Therefore, to react with 0.1 moles of H₂SO₄, we need:
\[ 0.1 \, \text{moles H}_2\text{SO}_4 \times 2 = 0.2 \, \text{moles NaOH} \]

Since we only have 0.1 moles of NaOH available, NaOH is the limiting reagent.

### Step 4: Calculate the moles of Na₂SO₄ produced
From the stoichiometry of the reaction:
- 2 moles of NaOH produce 1 mole of Na₂SO₄.

Thus, 0.1 moles of NaOH will produce:
\[ \text{Moles of Na}_2\text{SO}_4 = \frac{0.1 \, \text{moles NaOH}}{2} = 0.05 \, \text{moles Na}_2\text{SO}_4 \]

### Step 5: Calculate the mass of Na₂SO₄ formed
The molar mass of Na₂SO₄ can be calculated as follows:
- Na: 23 g/mol (2 Na = 46 g)
- S: 32 g/mol
- O: 16 g/mol (4 O = 64 g)

Total molar mass of Na₂SO₄:
\[ 46 + 32 + 64 = 142 \, \text{g/mol} \]

Now, calculate the mass of Na₂SO₄ formed:
\[ \text{Mass} = \text{Moles} \times \text{Molar Mass} \]
\[ \text{Mass of Na}_2\text{SO}_4 = 0.05 \, \text{moles} \times 142 \, \text{g/mol} = 7.1 \, \text{g} \]

### Step 6: Calculate the molarity of the remaining H₂SO₄
After the reaction, the moles of H₂SO₄ that reacted:
\[ \text{Moles of H}_2\text{SO}_4 \text{ reacted} = 0.05 \, \text{moles} \]
Remaining moles of H₂SO₄:
\[ \text{Remaining moles of H}_2\text{SO}_4 = 0.1 - 0.05 = 0.05 \, \text{moles} \]

The total volume of the solution after mixing is:
\[ \text{Total Volume} = 1 \, \text{L} + 1 \, \text{L} = 2 \, \text{L} \]

Now, calculate the molarity of the remaining H₂SO₄:
\[ \text{Molarity} = \frac{\text{Moles}}{\text{Volume (L)}} \]
\[ \text{Molarity of H}_2\text{SO}_4 = \frac{0.05 \, \text{moles}}{2 \, \text{L}} = 0.025 \, \text{M} \]

### Final Answers
1. Amount of sodium sulfate formed = 7.1 grams
2. Molarity of the solution = 0.025 M