Exactly 4.00 gm of a solution of `H_2 SO_4` was diluted with water and excess `BaC1_2` was added. The washed and dried precipitated of `BaSO_4` weighed 4.08 gm. The percent `H_2 SO_4` in the original acid solution is:

Options

A. `43.0%`
B. `4.3%`
C. `47%`
D. `56%`

Answer

To find the percentage of \( H_2SO_4 \) in the original acid solution, we can follow these steps:

### Step 1: Calculate the moles of \( BaSO_4 \)
We are given the mass of \( BaSO_4 \) precipitate, which is 4.08 g. We need to calculate the number of moles of \( BaSO_4 \) using the formula:

\[ \text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}} \]

The molar mass of \( BaSO_4 \) can be calculated as follows:
- Molar mass of \( Ba \) = 137 g/mol
- Molar mass of \( S \) = 32 g/mol
- Molar mass of \( O \) = 16 g/mol (there are 4 oxygen atoms)

\[ \text{Molar mass of } BaSO_4 = 137 + 32 + (4 \times 16) = 137 + 32 + 64 = 233 \text{ g/mol} \]

Now, substituting the values:

\[ \text{Moles of } BaSO_4 = \frac{4.08 \text{ g}}{233 \text{ g/mol}} \approx 0.0175 \text{ moles} \]

### Step 2: Relate moles of \( BaSO_4 \) to moles of \( H_2SO_4 \)
From the reaction, we know that 1 mole of \( H_2SO_4 \) produces 1 mole of \( BaSO_4 \). Therefore, the moles of \( H_2SO_4 \) will also be 0.0175 moles.

### Step 3: Calculate the mass of \( H_2SO_4 \)
To find the mass of \( H_2SO_4 \), we use its molar mass:

- Molar mass of \( H_2SO_4 \) = \( 2 \times 1 + 32 + (4 \times 16) = 2 + 32 + 64 = 98 \text{ g/mol} \)

Now, we can calculate the mass of \( H_2SO_4 \):

\[ \text{Mass of } H_2SO_4 = \text{Moles} \times \text{Molar Mass} = 0.0175 \text{ moles} \times 98 \text{ g/mol} \approx 1.715 \text{ g} \]

### Step 4: Calculate the percentage of \( H_2SO_4 \) in the original solution
The total mass of the solution is given as 4.00 g. The percentage of \( H_2SO_4 \) in the solution can be calculated as:

\[ \text{Percentage of } H_2SO_4 = \left( \frac{\text{Mass of } H_2SO_4}{\text{Mass of solution}} \right) \times 100 \]

Substituting the values:

\[ \text{Percentage of } H_2SO_4 = \left( \frac{1.715 \text{ g}}{4.00 \text{ g}} \right) \times 100 \approx 42.875\% \]

Rounding this to two decimal places gives us approximately 43%.

### Final Answer
The percentage of \( H_2SO_4 \) in the original acid solution is approximately **43%**.

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