This is a back titration question and I'm struggling with it :/ So if you can help me, that would be very appreciated :)
Q. A sample of steel wool was dissolved in excess sulfuric acid. The resulting solution (FeSO4 aq) was then titrated with potassium permanganate. 26.5mL of a 0.18 molL-1 of the permanganate solution was used in a titration with the solution. If the original mass of steel wood was 1.4g. Calculate the percentage of iron in the steel wool. NOTE: Fe2+ ions are oxidised to Fe3+ ions!
Thanks heaps!!
To calculate the percentage of iron in the steel wool, we need to determine the number of moles of iron and the mass of iron in the steel wool.
First, let's determine the number of moles of potassium permanganate used in the titration. We know the volume of the permanganate solution used and its concentration:
C1 = 0.18 mol/L (concentration of KMnO4 solution)
V1 = 26.5 mL (volume of KMnO4 solution used)
To convert the volume to liters, we divide by 1000:
V1 = 26.5 mL ÷ 1000 = 0.0265 L
Now, we can calculate the number of moles of permanganate used:
n1 = C1 × V1 = 0.18 mol/L × 0.0265 L = 0.00477 mol
Since the reaction between iron (Fe2+) and potassium permanganate (KMnO4) is a redox reaction, we can use stoichiometry to determine the number of moles of iron in the steel wool.
The balanced equation for the reaction is:
5Fe2+ + MnO4- + 8H+ → 5Fe3+ + Mn2+ + 4H2O
From the balanced equation, we know that 1 mole of permanganate reacts with 5 moles of Fe2+ ions.
Therefore, the number of moles of iron (Fe2+) in the steel wool is:
n2 = 5 × n1 = 5 × 0.00477 mol = 0.0239 mol
Next, let's calculate the mass of iron in the steel wool. We know the original mass of the steel wool, which is 1.4 g.
Now we need to calculate the molar mass of iron (Fe). The molar mass of Fe is approximately 55.85 g/mol.
Using the equation: mass = number of moles × molar mass, we can calculate the mass of iron:
mass of iron = n2 × molar mass of Fe = 0.0239 mol × 55.85 g/mol ≈ 1.34 g
Finally, calculate the percentage of iron in the steel wool:
percentage of iron = (mass of iron / original mass of steel wool) × 100
= (1.34 g / 1.4 g) × 100 ≈ 95.71%
Therefore, the percentage of iron in the steel wool is approximately 95.71%.