Calculate the number of millilitre of `NH_(3)`(aq) solution (d=0.986g/ml) contain 2.5% by mass `NH_(3)`, which will be required to precipitate iron as `Fe(OH)_(3)` in a 0.8 g sample that contains 50% `Fe_(2)O_(3)`.

Options

A. 0.344 mL
B. 3.44 mL
C. 17.24 mL
D. 10.34 mL

Answer

To solve the problem, we will follow these steps:

### Step 1: Calculate the mass of `Fe2O3` in the sample
Given that the sample weighs 0.8 g and contains 50% `Fe2O3`:
\[ \text{Mass of } Fe_2O_3 = 0.8 \, \text{g} \times 0.50 = 0.4 \, \text{g} \]

### Step 2: Calculate the moles of `Fe2O3`
The molar mass of `Fe2O3` is approximately 160 g/mol. We can calculate the moles of `Fe2O3` using the formula:
\[ \text{Moles of } Fe_2O_3 = \frac{\text{mass}}{\text{molar mass}} = \frac{0.4 \, \text{g}}{160 \, \text{g/mol}} = 2.5 \times 10^{-3} \, \text{moles} \]

### Step 3: Calculate the moles of `Fe3+`
Since each mole of `Fe2O3` produces 2 moles of `Fe3+`, we calculate:
\[ \text{Moles of } Fe^{3+} = 2 \times \text{Moles of } Fe_2O_3 = 2 \times 2.5 \times 10^{-3} = 5.0 \times 10^{-3} \, \text{moles} \]

### Step 4: Calculate the molarity of the `NH3` solution
The `NH3` solution has a density of 0.986 g/mL and is 2.5% by mass. First, we find the mass of `NH3` in 1 L of solution:
\[ \text{Mass of solution} = 0.986 \, \text{g/mL} \times 1000 \, \text{mL} = 986 \, \text{g} \]
\[ \text{Mass of } NH_3 = 0.025 \times 986 \, \text{g} = 24.65 \, \text{g} \]
The molar mass of `NH3` is approximately 17 g/mol, so:
\[ \text{Moles of } NH_3 = \frac{24.65 \, \text{g}}{17 \, \text{g/mol}} \approx 1.45 \, \text{moles} \]
Thus, the molarity of the `NH3` solution is:
\[ \text{Molarity} = \frac{\text{moles}}{\text{volume in L}} = \frac{1.45 \, \text{moles}}{1 \, \text{L}} = 1.45 \, \text{M} \]

### Step 5: Calculate the number of equivalents of `Fe3+`
The valency factor for `Fe3+` is 3, so:
\[ \text{Number of equivalents of } Fe^{3+} = \text{moles of } Fe^{3+} \times \text{valency factor} = 5.0 \times 10^{-3} \, \text{moles} \times 3 = 1.5 \times 10^{-2} \, \text{equivalents} \]

### Step 6: Calculate the volume of `NH3` solution required
Using the formula for equivalents:
\[ \text{Number of equivalents} = \text{Molarity} \times \text{Volume (L)} \]
We can rearrange this to find the volume:
\[ \text{Volume (L)} = \frac{\text{Number of equivalents}}{\text{Molarity}} = \frac{1.5 \times 10^{-2}}{1.45} \approx 0.01034 \, \text{L} \]
Converting to milliliters:
\[ \text{Volume (mL)} = 0.01034 \, \text{L} \times 1000 \, \text{mL/L} = 10.34 \, \text{mL} \]

### Final Answer
The volume of `NH3` solution required is approximately **10.34 mL**.

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