Calculate the mass of oxalic acid `(H_(2)C_(2)O_(4))` which can be oxidised to `CO_(2)` by `100.0mL` of `MnO_(4)^(-)` solution, `10 mL` of which is capable of oxidising `50.0mL` of `1.0NI^(-)` to `I_(2)`?

Options

A. 45g
B. 22.5g
C. 30g
D. 12.25g

Answer

To solve the problem of calculating the mass of oxalic acid (H₂C₂O₄) that can be oxidized to CO₂ by 100.0 mL of MnO₄⁻ solution, we will follow these steps:

### Step 1: Determine the Normality of the MnO₄⁻ Solution
From the problem, we know that 10 mL of the MnO₄⁻ solution can oxidize 50 mL of 1.0 N I⁻ to I₂.

Using the formula for equivalents:
\[ \text{Equivalents of I⁻} = \text{Normality} \times \text{Volume (L)} \]
\[ \text{Equivalents of I⁻} = 1.0 \, \text{N} \times 0.050 \, \text{L} = 0.050 \, \text{equivalents} \]

Since 10 mL of MnO₄⁻ solution oxidizes this amount of I⁻, the equivalents of MnO₄⁻ will also be 0.050 equivalents.

Now, we can find the normality of the MnO₄⁻ solution:
\[ \text{Normality of MnO₄⁻} = \frac{\text{Equivalents}}{\text{Volume (L)}} = \frac{0.050}{0.010} = 5.0 \, \text{N} \]

### Step 2: Calculate the Equivalents of MnO₄⁻ in 100 mL
Now, we need to find the equivalents of MnO₄⁻ in 100 mL:
\[ \text{Equivalents of MnO₄⁻} = \text{Normality} \times \text{Volume (L)} = 5.0 \, \text{N} \times 0.100 \, \text{L} = 0.50 \, \text{equivalents} \]

### Step 3: Write the Reaction for Oxidation of Oxalic Acid
The reaction for the oxidation of oxalic acid (H₂C₂O₄) by permanganate (MnO₄⁻) is:
\[ \text{MnO₄⁻} + \text{H₂C₂O₄} \rightarrow \text{Mn²⁺} + \text{CO₂} + \text{H₂O} \]

### Step 4: Determine the n-factor for Oxalic Acid
The n-factor for oxalic acid (H₂C₂O₄) is determined by the change in oxidation state. Each carbon in oxalic acid goes from +3 to +4 in CO₂, resulting in a total change of +1 for each carbon atom. Since there are two carbon atoms:
\[ \text{n-factor} = 2 \text{ (for 2 carbon atoms)} \]

### Step 5: Relate Equivalents of MnO₄⁻ to Oxalic Acid
Since the equivalents of MnO₄⁻ will equal the equivalents of oxalic acid:
\[ \text{Equivalents of H₂C₂O₄} = \text{Equivalents of MnO₄⁻} = 0.50 \, \text{equivalents} \]

Using the relationship:
\[ \text{Equivalents} = \text{Number of moles} \times \text{n-factor} \]
We can find the number of moles of oxalic acid:
\[ 0.50 = \text{Number of moles} \times 2 \implies \text{Number of moles} = \frac{0.50}{2} = 0.25 \, \text{moles} \]

### Step 6: Calculate the Mass of Oxalic Acid
Now, we can calculate the mass of oxalic acid using its molar mass (90 g/mol):
\[ \text{Mass} = \text{Number of moles} \times \text{Molar mass} = 0.25 \, \text{moles} \times 90 \, \text{g/mol} = 22.5 \, \text{g} \]

### Final Answer
The mass of oxalic acid that can be oxidized by 100.0 mL of MnO₄⁻ solution is **22.5 grams**.

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