a) To balance the equation, let's first determine the oxidation states of the involved elements. In the acidic solution, Fe is oxidized from Fe2+ to Fe3+ by KMnO4, which is reduced to Mn2+.
The balanced equation is as follows:
5Fe2+ + MnO4- + 8H+ → 5Fe3+ + Mn2+ + 4H2O
b) To calculate the percentage of iron in the sample, we need to find the number of moles of KMnO4 used in the titration and relate it to the number of moles of Fe in the sample.
Given:
Volume of KMnO4 solution used (V1) = 92.95 ml = 92.95/1000 L = 0.09295 L
Molarity of KMnO4 solution (M1) = 0.02 M
Using the balanced equation, we can determine the stoichiometric ratio between KMnO4 and Fe2+ as 1:5.
Thus, the number of moles of KMnO4 used (n1) can be calculated using the formula:
n1 = M1 × V1
Substituting the given values:
n1 = 0.02 M × 0.09295 L
Next, we know that 1 mole of KMnO4 reacts with 5 moles of Fe2+, so the number of moles of Fe2+ (n2) can be calculated as:
n2 = 5 × n1
Finally, to calculate the percentage of iron in the sample, we need to relate the mass of Fe to the number of moles of Fe and the molar mass of Fe.
Molar mass of Fe = 55.845 g/mol
Mass of Fe (m) = n2 × Molar mass of Fe
To find the percentage, we use the formula:
Percentage of Fe = (m / mass of the sample) × 100
The given information does not provide the mass of the sample, so we cannot calculate the exact percentage of iron without that information.