Final answer:
The percentage of iron in the ore sample can be calculated using the stoichiometry of the titration reaction. The mass percentage of iron in the ore sample is 25.73%. None of the given options match the calculated percentage.
Step-by-step explanation:
The percentage of iron in the ore sample can be calculated using the stoichiometry of the titration reaction. The balanced equation for the reaction between iron(II) and dichromate ion is:
6 Fe^2+ + Cr2O7^2- + 14 H+ → 6 Fe^3+ + 2 Cr^3+ + 7 H2O
From the balanced equation, we can see that 6 moles of Fe^2+ react with 1 mole of Cr2O7^2-. The number of moles of Cr2O7^2- used in the titration is calculated as:
moles of Cr2O7^2- = (volume of Na2Cr2O7 solution) × (concentration of Na2Cr2O7 solution) = (19.17 mL) × (0.0100 M) = 0.1917 moles
Since 6 moles of Fe^2+ react with 1 mole of Cr2O7^2-, the number of moles of Fe^2+ in the ore sample is calculated as:
moles of Fe^2+ = (moles of Cr2O7^2-) × (6 moles Fe^2+ / 1 mole Cr2O7^2-) = (0.1917 moles) × (6 moles Fe^2+ / 1 mole Cr2O7^2-) = 1.1502 moles
The mass percentage of iron in the ore sample is calculated as:
percentage of iron = (mass of Fe / mass of ore sample) × 100
The mass of Fe can be calculated using the molar mass of Fe:
mass of Fe = (moles of Fe^2+) × (molar mass of Fe) = (1.1502 moles) × (55.845 g/mol) = 64.317 g
The mass of the ore sample is given as 2.5000 g, so the percentage of iron in the sample is:
percentage of iron = (mass of Fe / mass of ore sample) × 100 = (64.317 g / 2.5000 g) × 100 = 25.73%
Therefore, the answer is None of the above, as none of the given options match the calculated percentage of iron in the ore sample.