How many grams of potassium dichromate are required to oxidise 20.0 g of `Fe^(2+)` in `FeSO_(4)` to `Fe^(3+)` if the reaction is carried out in an acidic medium ? (Molar mass of `K_(2)Cr_(21)O_(7)` and `FeSO_(4))` are 294 and 152 respectively

Explanation

To solve the problem of how many grams of potassium dichromate (K₂Cr₂O₇) are required to oxidize 20.0 g of Fe²⁺ in FeSO₄ to Fe³⁺ in an acidic medium, we will follow these steps:

### Step 1: Write the balanced redox reaction
The balanced reaction for the oxidation of Fe²⁺ to Fe³⁺ by potassium dichromate in acidic medium is:

\[ \text{K}_2\text{Cr}_2\text{O}_7 + 6 \text{FeSO}_4 + 14 \text{H}_2\text{SO}_4 \rightarrow 2 \text{Cr}_2\text{(SO}_4\text{)}_3 + 3 \text{Fe}_2\text{(SO}_4\text{)}_3 + 7 \text{H}_2\text{O} + K_2\text{SO}_4 \]

### Step 2: Calculate the moles of Fe²⁺
To find the moles of Fe²⁺ in 20.0 g of FeSO₄, we use the molar mass of FeSO₄, which is 152 g/mol.

\[ \text{Moles of Fe}^{2+} = \frac{\text{mass}}{\text{molar mass}} = \frac{20.0 \, \text{g}}{152 \, \text{g/mol}} \approx 0.1316 \, \text{mol} \]

### Step 3: Determine the stoichiometry of the reaction
From the balanced equation, we see that 1 mole of K₂Cr₂O₇ reacts with 6 moles of Fe²⁺. Therefore, to find the moles of K₂Cr₂O₇ needed, we use the stoichiometric ratio:

\[ \text{Moles of K}_2\text{Cr}_2\text{O}_7 = \frac{0.1316 \, \text{mol Fe}^{2+}}{6} \approx 0.02193 \, \text{mol K}_2\text{Cr}_2\text{O}_7 \]

### Step 4: Calculate the mass of K₂Cr₂O₇ required
Now, we can calculate the mass of K₂Cr₂O₇ needed using its molar mass (294 g/mol):

\[ \text{Mass of K}_2\text{Cr}_2\text{O}_7 = \text{moles} \times \text{molar mass} = 0.02193 \, \text{mol} \times 294 \, \text{g/mol} \approx 6.45 \, \text{g} \]

### Final Answer
The mass of potassium dichromate required to oxidize 20.0 g of Fe²⁺ in FeSO₄ to Fe³⁺ is approximately **6.45 grams**.

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