Exactly 4.00 gm of a solution of `H_2 SO_4` was diluted with water and excess `BaCl_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 solve the problem step by step, we will follow the chemical reaction and calculations as described in the video transcript.

### Step 1: Write the balanced chemical equation
The reaction between sulfuric acid (H₂SO₄) and barium chloride (BaCl₂) can be represented as:
\[ \text{H}_2\text{SO}_4 + \text{BaCl}_2 \rightarrow \text{BaSO}_4 + 2\text{HCl} \]

### Step 2: Calculate the moles of BaSO₄ formed
We know the mass of BaSO₄ precipitate is 4.08 g. To find the number of moles of BaSO₄, we use the formula:
\[ \text{Moles of BaSO}_4 = \frac{\text{mass of BaSO}_4}{\text{molar mass of BaSO}_4} \]

The molar mass of BaSO₄ (Barium Sulfate) is:
- Ba: 137.33 g/mol
- S: 32.07 g/mol
- O: 16.00 g/mol × 4 = 64.00 g/mol

Total molar mass of BaSO₄ = 137.33 + 32.07 + 64.00 = 233.40 g/mol

Now substituting the values:
\[ \text{Moles of BaSO}_4 = \frac{4.08 \, \text{g}}{233.40 \, \text{g/mol}} \approx 0.0175 \, \text{moles} \]

### Step 3: Relate moles of BaSO₄ to moles of H₂SO₄
From the balanced equation, we see that 1 mole of H₂SO₄ produces 1 mole of BaSO₄. Therefore, the moles of H₂SO₄ will also be:
\[ \text{Moles of H}_2\text{SO}_4 = 0.0175 \, \text{moles} \]

### Step 4: Calculate the mass of H₂SO₄ in the solution
To find the mass of H₂SO₄, we use its molar mass:
- Molar mass of H₂SO₄ = 2(1.01) + 32.07 + 4(16.00) = 98.09 g/mol

Now calculate the mass:
\[ \text{Mass of H}_2\text{SO}_4 = \text{Moles of H}_2\text{SO}_4 \times \text{Molar mass of H}_2\text{SO}_4 \]
\[ \text{Mass of H}_2\text{SO}_4 = 0.0175 \, \text{moles} \times 98.09 \, \text{g/mol} \approx 1.72 \, \text{g} \]

### Step 5: Calculate the percent concentration of H₂SO₄ in the solution
To find the percent concentration, we use the formula:
\[ \text{Percent H}_2\text{SO}_4 = \left( \frac{\text{Mass of H}_2\text{SO}_4}{\text{Mass of solution}} \right) \times 100 \]
\[ \text{Percent H}_2\text{SO}_4 = \left( \frac{1.72 \, \text{g}}{4.00 \, \text{g}} \right) \times 100 \approx 43.0\% \]

### Final Answer
The percent of H₂SO₄ in the original acid solution is approximately **43.0%**.

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