5 mol of `Fe_(2)(C_(2)O_(4))` is oxidised by `x mol` of `K_(2)Cr_(2)O_(7)` in acidic medium, calculate the value of `x`?

Answer

To solve the problem of how many moles of \( K_2Cr_2O_7 \) are required to oxidize 5 moles of \( Fe_2(C_2O_4) \) in acidic medium, we will follow these steps:

### Step 1: Write the oxidation half-reaction for \( Fe_2(C_2O_4) \)

The oxidation of iron in \( Fe_2(C_2O_4) \) can be represented as:
\[ Fe^{2+} \rightarrow Fe^{3+} + 3e^- \]
Since there are 2 moles of iron in \( Fe_2(C_2O_4) \), the total number of electrons released when 1 mole of \( Fe_2(C_2O_4) \) is oxidized is:
\[ 2 \times 3 = 6 \text{ electrons} \]

### Step 2: Write the reduction half-reaction for \( K_2Cr_2O_7 \)

In acidic medium, the reduction of dichromate ion \( Cr_2O_7^{2-} \) can be represented as:
\[ Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O \]
This shows that 1 mole of \( Cr_2O_7^{2-} \) accepts 6 electrons.

### Step 3: Establish the stoichiometric relationship

From the half-reactions, we see that:
- 2 moles of \( Fe_2(C_2O_4) \) release 6 electrons.
- 1 mole of \( K_2Cr_2O_7 \) consumes 6 electrons.

Thus, the balanced equation for the reaction is:
\[ 2 Fe_2(C_2O_4) + K_2Cr_2O_7 + 14H^+ \rightarrow 4Fe^{3+} + 2CO_2 + 7H_2O + 2Cr^{3+} \]

### Step 4: Calculate the moles of \( K_2Cr_2O_7 \) needed for 5 moles of \( Fe_2(C_2O_4) \)

From the balanced equation, we can see that 2 moles of \( Fe_2(C_2O_4) \) require 1 mole of \( K_2Cr_2O_7 \). Therefore, for 5 moles of \( Fe_2(C_2O_4) \):
\[ \text{Moles of } K_2Cr_2O_7 = \frac{5 \text{ moles of } Fe_2(C_2O_4)}{2} = \frac{5}{2} = 2.5 \text{ moles} \]

### Final Answer

Thus, the value of \( x \) is:
\[ x = 2.5 \]

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