`MnO_(2)` on ignition converts into `Mn_(3)O_(4)`. A sample of pyrolusite having 75% `MnO_(2)`, 20% inert impurities and rest water is ignited in air to constant mass. What is the percentage of Mn in the ignited sample ?

Options

A. (a) 0.246
B. (b) 0.37
C. (c) 0.5524
D. (d) 0.7405

Answer

To solve the problem step by step, we will follow these calculations:

### Step 1: Determine the composition of the sample
Given:
- 75% of the sample is `MnO2`
- 20% of the sample is inert impurities
- The rest (5%) is water

Assuming the total mass of the sample is `x` grams:
- Mass of `MnO2` = 75% of `x` = \(0.75x\)
- Mass of inert impurities = 20% of `x` = \(0.20x\)
- Mass of water = 5% of `x` = \(0.05x\)

### Step 2: Calculate the number of moles of `MnO2`
The molar mass of `MnO2` is calculated as follows:
- Atomic mass of Mn = 55 g/mol
- Atomic mass of O = 16 g/mol
- Molar mass of `MnO2` = \(55 + 2 \times 16 = 87 \, \text{g/mol}\)

Now, calculate the number of moles of `MnO2`:
\[ \text{Number of moles of } MnO2 = \frac{\text{mass of } MnO2}{\text{molar mass of } MnO2} = \frac{0.75x}{87} \]

### Step 3: Calculate the mass of manganese (Mn)
Since each mole of `MnO2` contains one mole of Mn, the number of moles of Mn will be the same as that of `MnO2`:
\[ \text{Number of moles of Mn} = \frac{0.75x}{87} \]

Now, calculate the mass of Mn:
\[ \text{Mass of Mn} = \text{Number of moles of Mn} \times \text{atomic mass of Mn} = \left(\frac{0.75x}{87}\right) \times 55 = \frac{0.474x}{1} \]

### Step 4: Determine the reaction during ignition
The reaction during ignition is:
\[ 3MnO2 \rightarrow Mn3O4 + O2 \]
From the stoichiometry of the reaction, 3 moles of `MnO2` yield 1 mole of `Mn3O4`.

### Step 5: Calculate the number of moles of `Mn3O4`
The number of moles of `Mn3O4` produced is:
\[ \text{Number of moles of } Mn3O4 = \frac{1}{3} \times \text{Number of moles of } MnO2 = \frac{1}{3} \times \frac{0.75x}{87} = \frac{0.25x}{87} \]

### Step 6: Calculate the mass of `Mn3O4`
The molar mass of `Mn3O4` is:
- Molar mass of `Mn3O4` = \(3 \times 55 + 4 \times 16 = 229 \, \text{g/mol}\)

Now, calculate the mass of `Mn3O4`:
\[ \text{Mass of } Mn3O4 = \text{Number of moles of } Mn3O4 \times \text{molar mass of } Mn3O4 = \left(\frac{0.25x}{87}\right) \times 229 = \frac{0.644x}{1} \]

### Step 7: Calculate the total weight of the ignited sample
The total weight of the ignited sample (residue) will be the sum of the mass of `Mn3O4` and the mass of inert impurities:
\[ \text{Total weight of residue} = \text{Mass of } Mn3O4 + \text{Mass of inert impurities} = \frac{0.644x}{1} + 0.20x = 0.844x \]

### Step 8: Calculate the percentage of Mn in the ignited sample
Now, we can find the percentage of Mn in the ignited sample:
\[ \text{Percentage of Mn} = \left(\frac{\text{Mass of Mn}}{\text{Total weight of residue}}\right) \times 100 = \left(\frac{0.474x}{0.844x}\right) \times 100 \]

Simplifying this gives:
\[ \text{Percentage of Mn} = \left(\frac{0.474}{0.844}\right) \times 100 \approx 56.05\% \]

### Final Result
Thus, the percentage of Mn in the ignited sample is approximately **56.05%**.

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