In the balanced equation
`MnO_(4)^(-)+H^(+)+C_(2)O_(4)^(2-)rarr Mn^(2+)+CO_(2)+H_(2)O`, the moles of `CO_(2)` formed are :-

Options

A. 2
B. 4
C. 5
D. 10

Answer

To solve the question regarding the balanced equation

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

we need to determine the moles of \(\text{CO}_2\) formed. We will do this by balancing the redox reaction step by step.

### Step 1: Identify the half-reactions
The first step is to separate the reaction into oxidation and reduction half-reactions.

- **Reduction half-reaction**: The manganese in \(\text{MnO}_4^{-}\) is reduced to \(\text{Mn}^{2+}\).
- **Oxidation half-reaction**: The oxalate ion \(\text{C}_2\text{O}_4^{2-}\) is oxidized to \(\text{CO}_2\).

### Step 2: Write the reduction half-reaction
The reduction half-reaction for manganese can be written as:

\[ \text{MnO}_4^{-} + 8\text{H}^{+} + 5\text{e}^{-} \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} \]

### Step 3: Write the oxidation half-reaction
The oxidation half-reaction for oxalate can be written as:

\[ \text{C}_2\text{O}_4^{2-} \rightarrow 2\text{CO}_2 + 2\text{e}^{-} \]

### Step 4: Balance the electrons
To balance the overall reaction, we need to ensure that the number of electrons lost in the oxidation half-reaction equals the number of electrons gained in the reduction half-reaction.

The reduction half-reaction involves 5 electrons, while the oxidation half-reaction involves 2 electrons. To balance, we can multiply the oxidation half-reaction by 5 and the reduction half-reaction by 2:

- **Reduction half-reaction** (multiplied by 2):
\[ 2\text{MnO}_4^{-} + 16\text{H}^{+} + 10\text{e}^{-} \rightarrow 2\text{Mn}^{2+} + 8\text{H}_2\text{O} \]

- **Oxidation half-reaction** (multiplied by 5):
\[ 5\text{C}_2\text{O}_4^{2-} \rightarrow 10\text{CO}_2 + 10\text{e}^{-} \]

### Step 5: Combine the half-reactions
Now, we can combine both half-reactions:

\[ 2\text{MnO}_4^{-} + 16\text{H}^{+} + 5\text{C}_2\text{O}_4^{2-} \rightarrow 2\text{Mn}^{2+} + 10\text{CO}_2 + 8\text{H}_2\text{O} \]

### Step 6: Determine the moles of \(\text{CO}_2\)
From the balanced equation, we can see that for every 5 moles of \(\text{C}_2\text{O}_4^{2-}\) that react, 10 moles of \(\text{CO}_2\) are produced.

Thus, the moles of \(\text{CO}_2\) formed are **10 moles**.

### Summary
The balanced reaction shows that 10 moles of \(\text{CO}_2\) are produced when the reaction occurs.

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