How many grams of `K_(2)Cr_(2)O_(7)` are required to oxidize `Fe^(2+)` present in 15.2 g of `FeSO_(4) to Fe^(3+)` if the reaction is carried out in an acidic medium ?

Explanation

To solve the problem of how many grams of \( K_2Cr_2O_7 \) are required to oxidize \( Fe^{2+} \) present in 15.2 g of \( FeSO_4 \) to \( Fe^{3+} \) in an acidic medium, we can follow these steps:

### Step 1: Calculate the moles of \( FeSO_4 \)
First, we need to find the number of moles of \( FeSO_4 \) in 15.2 g.

- The molar mass of \( FeSO_4 \) can be calculated as follows:
- Molar mass of \( Fe \) = 55.85 g/mol
- Molar mass of \( S \) = 32.07 g/mol
- Molar mass of \( O_4 \) = 4 × 16.00 g/mol = 64.00 g/mol
- Total molar mass of \( FeSO_4 \) = 55.85 + 32.07 + 64.00 = 151.92 g/mol

Now, calculate the moles of \( FeSO_4 \):
\[ \text{Moles of } FeSO_4 = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{15.2 \, \text{g}}{151.92 \, \text{g/mol}} \approx 0.100 \, \text{mol} \]

### Step 2: Determine the moles of \( Fe^{2+} \)
Since each mole of \( FeSO_4 \) contains one mole of \( Fe^{2+} \), the moles of \( Fe^{2+} \) will also be 0.100 mol.

### Step 3: Calculate the equivalent of \( Fe^{2+} \)
The oxidation of \( Fe^{2+} \) to \( Fe^{3+} \) involves a change of one electron. Therefore, the number of equivalents of \( Fe^{2+} \) is equal to the number of moles:
\[ \text{Equivalents of } Fe^{2+} = \text{Moles of } Fe^{2+} = 0.100 \, \text{eq} \]

### Step 4: Determine the equivalents of \( K_2Cr_2O_7 \)
In acidic medium, \( K_2Cr_2O_7 \) is reduced from \( Cr^{6+} \) to \( Cr^{3+} \), which involves a change of 6 electrons (as each \( Cr \) changes from +6 to +3). Therefore, the equivalent factor (n-factor) for \( K_2Cr_2O_7 \) is 6.

### Step 5: Calculate the required equivalents of \( K_2Cr_2O_7 \)
To find the equivalents of \( K_2Cr_2O_7 \) needed to oxidize \( Fe^{2+} \):
\[ \text{Equivalents of } K_2Cr_2O_7 = \frac{\text{Equivalents of } Fe^{2+}}{\text{n-factor of } K_2Cr_2O_7} = \frac{0.100 \, \text{eq}}{6} \approx 0.01667 \, \text{eq} \]

### Step 6: Calculate the moles of \( K_2Cr_2O_7 \)
Using the equivalents and the n-factor:
\[ \text{Moles of } K_2Cr_2O_7 = \frac{\text{Equivalents}}{\text{n-factor}} = \frac{0.01667 \, \text{eq}}{1} \approx 0.01667 \, \text{mol} \]

### Step 7: Calculate the mass of \( K_2Cr_2O_7 \)
Now, we can calculate the mass of \( K_2Cr_2O_7 \) required using its molar mass.

- Molar mass of \( K_2Cr_2O_7 \):
- Molar mass of \( K \) = 39.10 g/mol (2 K) = 78.20 g/mol
- Molar mass of \( Cr \) = 51.996 g/mol (2 Cr) = 103.992 g/mol
- Molar mass of \( O_7 \) = 7 × 16.00 g/mol = 112.00 g/mol
- Total molar mass of \( K_2Cr_2O_7 \) = 78.20 + 103.992 + 112.00 = 294.192 g/mol

Now, calculate the mass:
\[ \text{Mass of } K_2Cr_2O_7 = \text{Moles} \times \text{Molar Mass} = 0.01667 \, \text{mol} \times 294.192 \, \text{g/mol} \approx 4.91 \, \text{g} \]

### Final Answer
Thus, approximately 4.91 grams of \( K_2Cr_2O_7 \) are required to oxidize \( Fe^{2+} \) present in 15.2 g of \( FeSO_4 \) to \( Fe^{3+} \) in an acidic medium.

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