Question

The value of x is ________.

Answer 1

Correct Answer: 1.875

Step 1: Given reaction
The given chemical reaction is:
\( 8H^+ + 5Fe^{2+} + MnO_4^- \rightarrow 5Fe^{3+} + Mn^{2+} + 4H_2O \)

Step 2: Calculation of meq of Fe²⁺ and MnO^4– for 25 ml
We are given the following relation:
meq of Fe²⁺ = meq of MnO^4–.
For 25 ml, the meq is calculated as:
\( \text{meq of Fe}^{2+} = 12.5 \times 0.03 \times 5 \).

Step 3: Calculation for 250 ml
For 250 ml, we use the formula to calculate the moles of Fe²⁺ as follows:
\( \text{moles of Fe}^{2+} = 12.5 \times 0.03 \times 5 \times \frac{250}{25} \)
This simplifies to:
\( \text{moles of Fe}^{2+} = \frac{18.75}{1000} \) mol.

Step 4: Final Calculation
Thus, the moles of Fe²⁺ are:
\( \text{moles of Fe}^{2+} = 18.75 \times 10^{-3} \) mol.
Simplified further:
\( \text{moles of Fe}^{2+} = 1.875 \times 10^{-2} \) mol.

Final Answer:
\( x = 1.875 \).

Answer 2

Step 1: Given data
- Mass of sample = 5.6 g
- Volume of solution = 250 mL
- Volume of 25.0 mL of solution used for titration
- Volume of KMnO₄ used = 12.5 mL
- Molarity of KMnO₄ = 0.03 M
- Molar mass of iron = 56 g/mol
- KMnO₄ reacts only with Fe²⁺ in the solution.

Step 2: Write the balanced reaction
The reaction between KMnO₄ and Fe²⁺ is given by:
\( 5Fe^{2+} + MnO_4^- + 8H^+ \rightarrow 5Fe^{3+} + Mn^{2+} + 4H_2O \)
This means that 5 moles of Fe²⁺ react with 1 mole of MnO₄.

Step 3: Calculate the number of moles of KMnO₄ used in titration
We are given the volume and molarity of KMnO₄ used in titration.
The moles of KMnO₄ are calculated using the formula:
\( \text{moles of KMnO}_4 = Molarity \times Volume \)
\( = 0.03 \, \text{mol/L} \times 12.5 \times 10^{-3} \, \text{L} \)
\( = 3.75 \times 10^{-4} \, \text{mol} \)

Step 4: Calculate the number of moles of Fe²⁺ reacting with KMnO₄
From the balanced reaction, 5 moles of Fe²⁺ react with 1 mole of KMnO₄.
So, the moles of Fe²⁺ will be 5 times the moles of KMnO₄.
Therefore, the moles of Fe²⁺ reacting with 3.75 × 10^–4 moles of KMnO₄ are:
\( \text{moles of Fe}^{2+} = 5 \times 3.75 \times 10^{-4} \)
\( = 1.875 \times 10^{-3} \, \text{mol} \)

Step 5: Calculate the moles of Fe²⁺ in 250 mL solution
The moles of Fe²⁺ in 25 mL solution is 1.875 × 10^–3 mol.
To find the moles of Fe²⁺ in the entire 250 mL solution, we multiply by the dilution factor:
\( \text{moles of Fe}^{2+} = 1.875 \times 10^{-3} \times \frac{250}{25} \)
\( = 1.875 \times 10^{-3} \times 10 = 1.875 \times 10^{-2} \, \text{mol} \)

Step 6: Calculate the mass of Fe in the sample
The number of moles of Fe²⁺ in 250 mL solution is 1.875 × 10^–2 mol.
Since 1 mole of Fe²⁺ corresponds to 56 g of iron, the mass of iron in the sample is:
\( \text{mass of Fe} = 1.875 \times 10^{-2} \times 56 \)
\( = 1.05 \, \text{g} \)

Step 7: Calculate the percentage of iron (y%) in the sample
The mass of the sample is 5.6 g, and the mass of iron is 1.05 g.
The percentage of iron in the sample is calculated as:
\( y\% = \frac{\text{mass of Fe}}{\text{mass of sample}} \times 100 \)
\( y\% = \frac{1.05}{5.6} \times 100 \)
\( y\% = 18.75\% \)

Final Answer:
The percentage of iron in the sample is 18.75%.