Find the number of moles of `K_(2)Cr_(2)O_(7)` reduced by three mole of `Sn^(2+)` ions.

Explanation

To find the number of moles of \( K_2Cr_2O_7 \) reduced by three moles of \( Sn^{2+} \) ions, we can follow these steps:

### Step 1: Write the half-reaction for the reduction of \( K_2Cr_2O_7 \)
The dichromate ion \( Cr_2O_7^{2-} \) is reduced to \( Cr^{3+} \). The half-reaction can be written as:
\[ Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O \]
This indicates that 1 mole of \( Cr_2O_7^{2-} \) requires 6 moles of electrons for reduction.

### Step 2: Write the half-reaction for the oxidation of \( Sn^{2+} \)
The oxidation of \( Sn^{2+} \) to \( Sn^{4+} \) can be written as:
\[ Sn^{2+} \rightarrow Sn^{4+} + 2e^- \]
This indicates that 1 mole of \( Sn^{2+} \) produces 2 moles of electrons.

### Step 3: Calculate the total moles of electrons from \( Sn^{2+} \)
Given that we have 3 moles of \( Sn^{2+} \):
\[ 3 \, moles \, of \, Sn^{2+} \times 2 \, moles \, of \, e^- \, per \, mole \, of \, Sn^{2+} = 6 \, moles \, of \, e^- \]

### Step 4: Relate the moles of electrons to moles of \( K_2Cr_2O_7 \)
From the half-reaction of \( Cr_2O_7^{2-} \), we know that 1 mole of \( Cr_2O_7^{2-} \) requires 6 moles of electrons. Therefore, the number of moles of \( K_2Cr_2O_7 \) that can be reduced by 6 moles of electrons is:
\[ \frac{6 \, moles \, of \, e^-}{6 \, moles \, of \, e^- \, per \, mole \, of \, Cr_2O_7^{2-}} = 1 \, mole \, of \, K_2Cr_2O_7 \]

### Conclusion
Thus, the number of moles of \( K_2Cr_2O_7 \) reduced by three moles of \( Sn^{2+} \) ions is **1 mole**.

---