Iodobenzene `(C_(6)H_(5)l)` is prepared from aniline `(C_(6)H_(5)NH_(2))` in a two step process as shown below `C_(6)H_(5)NH_(2) + HNO_(2) + HCl rarr C_(6)H_(5)N_(2) .^(+)Cl^(-) + 2H_(2)O " "C_(6)H_(5)N_(2) .^(+)Cl^(-) + KI rarr C_(6)H_(5)I + N_(2) + KCl`
In an actual preparation `9.30 g` of aniline was coverted to `16.32 g` of iodobenzene. The percentage yield of iodobenzene is :

Options

A. `8%`
B. `50%`
C. `75%`
D. `80%`

Explanation

To calculate the percentage yield of iodobenzene from aniline, we will follow these steps:

### Step 1: Calculate the molar mass of aniline (C₆H₅NH₂) and iodobenzene (C₆H₅I)
- Molar mass of aniline (C₆H₅NH₂):
- Carbon (C): 12.01 g/mol × 6 = 72.06 g/mol
- Hydrogen (H): 1.008 g/mol × 7 = 7.056 g/mol (5 from C₆H₅ and 2 from NH₂)
- Nitrogen (N): 14.01 g/mol × 1 = 14.01 g/mol
- Total = 72.06 + 7.056 + 14.01 = 93.126 g/mol (approximately 93.13 g/mol)

- Molar mass of iodobenzene (C₆H₅I):
- Carbon (C): 12.01 g/mol × 6 = 72.06 g/mol
- Hydrogen (H): 1.008 g/mol × 5 = 5.04 g/mol
- Iodine (I): 126.90 g/mol × 1 = 126.90 g/mol
- Total = 72.06 + 5.04 + 126.90 = 203.00 g/mol (approximately 204 g/mol)

### Step 2: Calculate the theoretical yield of iodobenzene from the given mass of aniline
- Given mass of aniline = 9.30 g
- Moles of aniline = mass / molar mass = 9.30 g / 93.13 g/mol = 0.0998 mol (approximately)

- According to the reaction, 1 mole of aniline produces 1 mole of iodobenzene.
- Therefore, moles of iodobenzene produced = moles of aniline = 0.0998 mol

- Theoretical mass of iodobenzene = moles × molar mass of iodobenzene
= 0.0998 mol × 204 g/mol = 20.39 g (approximately)

### Step 3: Calculate the percentage yield of iodobenzene
- Actual yield of iodobenzene = 16.32 g
- Percentage yield = (actual yield / theoretical yield) × 100
= (16.32 g / 20.39 g) × 100 = 80.00% (approximately)

### Final Answer
The percentage yield of iodobenzene is approximately **80%**.

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