The problem asks for the stoichiometric ratio of Mohr's salt to potassium dichromate in a redox titration. We need to find the balanced chemical equation for the reaction.

Step 1: Identify the reactants and their roles.

• Potassium dichromate ($K_2Cr_2O_7$): The active ion is the dichromate ion, $Cr_2O_7^{2-}$. In acidic medium, it is a strong oxidizing agent.
• Mohr's salt ($FeSO_4 \cdot (NH_4)_2SO_4 \cdot 6H_2O$): The active ion is the ferrous ion, $Fe^{2+}$. It acts as the reducing agent.
• The reaction is carried out in an acidified solution, so $H^+$ ions are available.

Step 2: Write the half-reactions for oxidation and reduction.

• Oxidation half-reaction: The ferrous ion ($Fe^{2+}$) is oxidized to the ferric ion ($Fe^{3+}$). $Fe^{2+} \rightarrow Fe^{3+} + e^-$

• Reduction half-reaction: The dichromate ion ($Cr_2O_7^{2-}$) is reduced to the chromium(III) ion ($Cr^{3+}$) in acidic medium.

  1. Start with the core species: $Cr_2O_7^{2-} \rightarrow Cr^{3+}$
  2. Balance the Chromium (Cr) atoms: $Cr_2O_7^{2-} \rightarrow 2Cr^{3+}$
  3. Balance the Oxygen (O) atoms by adding water ($H_2O$): $Cr_2O_7^{2-} \rightarrow 2Cr^{3+} + 7H_2O$
  4. Balance the Hydrogen (H) atoms by adding protons ($H^+$): $Cr_2O_7^{2-} + 14H^+ \rightarrow 2Cr^{3+} + 7H_2O$
  5. Balance the charge by adding electrons ($e^-$). The charge on the left is $(-2) + (+14) = +12$. The charge on the right is $2 \times (+3) = +6$. To balance, we add 6 electrons to the left side. $Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$

Step 3: Combine the half-reactions to get the overall balanced equation.

To combine the two half-reactions, the number of electrons lost in oxidation must equal the number of electrons gained in reduction.

• Oxidation: $Fe^{2+} \rightarrow Fe^{3+} + e^-$ (1 electron)
• Reduction: $Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$ (6 electrons)

We need to multiply the oxidation half-reaction by 6 to balance the electrons: $6Fe^{2+} \rightarrow 6Fe^{3+} + 6e^-$

Now, add the modified oxidation half-reaction and the reduction half-reaction: $6Fe^{2+} + Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 6Fe^{3+} + 2Cr^{3+} + 7H_2O + 6e^-$

Cancel the electrons ($6e^-$) from both sides to get the final balanced ionic equation: $6Fe^{2+} + Cr_2O_7^{2-} + 14H^+ \rightarrow 6Fe^{3+} + 2Cr^{3+} + 7H_2O$

Step 4: Determine the molar ratio.

From the balanced equation, we can see that 6 moles of $Fe^{2+}$ ions react with 1 mole of $Cr_2O_7^{2-}$ ions. Since one mole of Mohr's salt provides one mole of $Fe^{2+}$, it follows that 6 moles of Mohr's salt are required to react with 1 mole of potassium dichromate.

Alternative Method (Using n-factor/equivalent concept):

• The n-factor of a substance in a redox reaction is the number of moles of electrons lost or gained per mole of the substance.
• n-factor of Mohr's salt ($Fe^{2+} \rightarrow Fe^{3+}$): The change in oxidation state of iron is from +2 to +3. The change is 1. So, n-factor = 1.
• n-factor of Potassium Dichromate ($K_2Cr_2O_7$): The oxidation state of Cr changes from +6 in $Cr_2O_7^{2-}$ to +3 in $Cr^{3+}$. The change per Cr atom is 3. Since there are two Cr atoms in $K_2Cr_2O_7$, the total change is $2 \times 3 = 6$. So, n-factor = 6.

According to the law of equivalence, at the equivalence point: Equivalents of Mohr's salt = Equivalents of Potassium Dichromate $(moles \times n-factor)_{Mohr's\ salt} = (moles \times n-factor)_{K_2Cr_2O_7}$ $n_{Mohr's\ salt} \times 1 = n_{K_2Cr_2O_7} \times 6$

The question asks for the number of moles of Mohr's salt required per mole of dichromate, which is the ratio $\frac{n_{Mohr's\ salt}}{n_{K_2Cr_2O_7}}$. $\frac{n_{Mohr's\ salt}}{n_{K_2Cr_2O_7}} = \frac{6}{1} = 6$

Both methods show that 6 moles of Mohr's salt are required per mole of potassium dichromate.

Conclusion:

Comparing our result with the options: A: 3 B: 4 C: 5 D: 6

The correct option is D.