The number of moles of `K_(2)Cr_(2)O_(7)` that will be needed to react completely with one mole of ferric sulphite in acidic medium is

Options

A. `0.4`
B. `0.6`
C. `1.0`
D. `0.8`

Explanation

To determine the number of moles of \( K_2Cr_2O_7 \) needed to react completely with one mole of ferric sulfite (\( Fe_2(SO_3)_3 \)) in an acidic medium, we will follow these steps:

### Step 1: Write the balanced chemical equation
In acidic medium, potassium dichromate (\( K_2Cr_2O_7 \)) acts as an oxidizing agent and ferric sulfite (\( Fe_2(SO_3)_3 \)) is oxidized. The half-reactions can be written as follows:

1. The reduction half-reaction for dichromate:
\[ Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O \]

2. The oxidation half-reaction for ferric sulfite:
\[ 6Fe^{2+} \rightarrow 6Fe^{3+} + 6e^- \]

### Step 2: Combine the half-reactions
To combine these half-reactions, 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. From the equations, we see that 6 moles of \( Fe^{2+} \) will produce 6 moles of electrons, which matches the 6 moles of electrons required by the dichromate.

The overall balanced reaction will be:
\[ Cr_2O_7^{2-} + 6Fe^{2+} + 14H^+ \rightarrow 2Cr^{3+} + 6Fe^{3+} + 7H_2O \]

### Step 3: Determine the mole ratio
From the balanced equation, we can see that:
- 1 mole of \( K_2Cr_2O_7 \) reacts with 6 moles of \( Fe^{2+} \).

### Step 4: Calculate the moles of \( K_2Cr_2O_7 \) needed for 1 mole of \( Fe^{2+} \)
Since 1 mole of \( K_2Cr_2O_7 \) reacts with 6 moles of \( Fe^{2+} \), we can find out how much \( K_2Cr_2O_7 \) is needed for 1 mole of \( Fe^{2+} \):
\[ \text{Moles of } K_2Cr_2O_7 = \frac{1 \text{ mole of } Fe^{2+}}{6} = \frac{1}{6} \text{ moles of } K_2Cr_2O_7 \]

### Step 5: Final answer
Thus, the number of moles of \( K_2Cr_2O_7 \) needed to react completely with 1 mole of ferric sulfite in acidic medium is:
\[ \frac{1}{6} \text{ moles or approximately } 0.17 \text{ moles} \]