Iron(II) sulfate, is typically the source of iron in tablets prescribed to anemic patients, a condition where a person lacks enough healthy red blood cells. The tablets can be dissolved in dilute sulfuric acid to dissolve the Fe2+ ions. To determine the percentage by mass of iron within the tablets, the Fe2+ ions can be oxidised to Fe3+ using potassium permanganate, where MnO4− is reduced to Mn2+. Calculate the percentage by mass of iron in a 500 mg tablet that requires 24.53 cm3 of 0.0100 mol dm−3 KMnO4 to titrate the iron sample. State your answer to 3 significant figures. M(Fe) = 55.85 g mol−1
Answers
Answer:
The percentage by mass of iron in the tablet is 54.9%, rounded to 3 significant figures.
Explanation:
The balanced chemical equation for the oxidation of Fe2+ to Fe3+ using KMnO4 can be written as:
5Fe2+ + MnO4- + 8H+ -> 5Fe3+ + Mn2+ + 4H2O
From the balanced equation, we can see that 1 mole of KMnO4 reacts with 5 moles of Fe2+ to produce 5 moles of Fe3+. Therefore, we can calculate the number of moles of Fe2+ in the tablet using the volume and concentration of KMnO4 used in the titration:
Number of moles of KMnO4 = concentration x volume
Number of moles of Fe2+ = (1/5) x number of moles of KMnO4
Number of moles of KMnO4 = 0.0100 mol dm−3 x 0.02453 dm3
Number of moles of KMnO4 = 0.0002453 mol
Number of moles of Fe2+ = (1/5) x 0.0002453 mol
Number of moles of Fe2+ = 0.00004906 mol
The mass of iron in the tablet can be calculated using the number of moles of Fe2+ and the molar mass of iron:
Mass of iron = number of moles of Fe2+ x molar mass of Fe
Mass of iron = 0.00004906 mol x 55.85 g/mol
Mass of iron = 0.002743 g
Finally, we can calculate the percentage by mass of iron in the tablet:
Percentage by mass of iron = (mass of iron / mass of tablet) x 100%
Percentage by mass of iron = (0.002743 g / 0.5 g) x 100%
Percentage by mass of iron = 0.5486 x 100%
Percentage by mass of iron = 54.9%
Therefore, the percentage by mass of iron in the tablet is 54.9%, rounded to 3 significant figures.
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