15 mol of `KMnO_(4)` are treated with excess `H_(2)C_(2)O_(4)` in `H_(2)SO_(4)` medium. How many moles of `CO_(2)` will be formed and how many moles of `H_(2)C_(2)O_(4)` will be consumed?

Options

A. 75,37.5
B. 3,15
C. 3,6
D. 75,150

Answer

To solve the problem of how many moles of \( CO_2 \) will be formed and how many moles of \( H_2C_2O_4 \) will be consumed when 15 moles of \( KMnO_4 \) are treated with excess \( H_2C_2O_4 \) in \( H_2SO_4 \) medium, we will follow these steps:

### Step 1: Write the balanced redox reaction
The reaction between potassium permanganate (\( KMnO_4 \)) and oxalic acid (\( H_2C_2O_4 \)) in acidic medium can be represented as follows:

\[ \text{MnO}_4^- + \text{C}_2O_4^{2-} \rightarrow \text{Mn}^{2+} + \text{CO}_2 + \text{H}_2O \]

### Step 2: Determine the oxidation states
- In \( KMnO_4 \), manganese (Mn) is in the +7 oxidation state.
- In \( C_2O_4^{2-} \), carbon (C) is in the +3 oxidation state.
- In \( CO_2 \), carbon (C) is in the +4 oxidation state.
- Therefore, Mn is reduced from +7 to +2, and C is oxidized from +3 to +4.

### Step 3: Balance the half-reactions
- For the reduction half-reaction:
\[ \text{MnO}_4^- + 8H^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4H_2O \]
- For the oxidation half-reaction:
\[ \text{C}_2O_4^{2-} \rightarrow 2CO_2 + 2e^- \]

### Step 4: Combine the half-reactions
To balance the electrons transferred, we multiply the oxidation half-reaction by 5:
\[ 5 \text{C}_2O_4^{2-} \rightarrow 10CO_2 + 10e^- \]

Now we can combine the two half-reactions:
\[ 2 \text{MnO}_4^- + 5 \text{C}_2O_4^{2-} + 16H^+ \rightarrow 2 \text{Mn}^{2+} + 10CO_2 + 8H_2O \]

### Step 5: Determine the mole ratios
From the balanced equation:
- 2 moles of \( KMnO_4 \) produce 10 moles of \( CO_2 \).
- 2 moles of \( KMnO_4 \) consume 5 moles of \( C_2O_4^{2-} \).

### Step 6: Calculate moles of \( CO_2 \) and \( H_2C_2O_4 \)
Given that we have 15 moles of \( KMnO_4 \):
- Using the mole ratio for \( CO_2 \):
\[ \text{Moles of } CO_2 = 15 \text{ moles } KMnO_4 \times \frac{10 \text{ moles } CO_2}{2 \text{ moles } KMnO_4} = 75 \text{ moles } CO_2 \]

- Using the mole ratio for \( H_2C_2O_4 \):
\[ \text{Moles of } H_2C_2O_4 = 15 \text{ moles } KMnO_4 \times \frac{5 \text{ moles } C_2O_4^{2-}}{2 \text{ moles } KMnO_4} = 37.5 \text{ moles } H_2C_2O_4 \]

### Final Answer
- **Moles of \( CO_2 \) formed:** 75 moles
- **Moles of \( H_2C_2O_4 \) consumed:** 37.5 moles

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