A 1.362 g sample of an iron ore that contained Fe_3O_4 was dissolved in acid and all the iron was reduced to Fe^2+. the solution was then acidified with H_2SO_4 and titrated with 39.42 mL of 0.0281 M KMnO_4 , which oxidized the iron to Fe^3+.
Net ionic equation
5Fe^(2+)+MnO_4^-+8H^+→5Fe^(3+)+Mn^(2+)+4H_2O
a) what was the percentage by mass of iron in the ore?
b) what was the percentage by mass of Fe_3O_5 in the or
There is a chemist around but how would he know to read this?
To determine the percentage by mass of iron in the ore and the percentage by mass of Fe3O5, we need to use stoichiometry and the concept of molar ratios.
a) Percentage by mass of iron in the ore:
1. Calculate the number of moles of Fe2+ ions present in the solution.
Since 1 mole of Fe2+ ions reacts with 1 mole of MnO4-, the moles of Fe2+ ions can be calculated using the molar concentration of KMnO4 and the volume of KMnO4 used in the titration.
Moles of Fe2+ = molar concentration of KMnO4 * volume of KMnO4 used (in liters)
= 0.0281 M * (39.42 mL / 1000 mL/L)
= 0.001107942 moles of Fe2+
2. Use the stoichiometric coefficient in the balanced equation to determine the moles of Fe3+ ions produced.
According to the balanced equation, 5 moles of Fe2+ ions react to produce 5 moles of Fe3+ ions.
Moles of Fe3+ = 0.001107942 moles of Fe2+
= 0.001107942 moles of Fe3+
3. Calculate the molar mass of iron (Fe).
Molar mass of Fe = atomic mass of Fe
= 55.845 g/mol
4. Calculate the mass of iron in the ore.
Mass of iron = Moles of Fe3+ * Molar mass of Fe
= 0.001107942 moles * 55.845 g/mol
= 0.06195317 g
5. Calculate the percentage by mass of iron in the ore.
Percentage by mass of iron = (Mass of iron / Mass of the ore sample) * 100
= (0.06195317 g / 1.362 g) * 100
= 4.55%
Therefore, the percentage by mass of iron in the ore is approximately 4.55%.
b) Percentage by mass of Fe3O5 in the ore:
To determine the percentage by mass of Fe3O5, we need to assume that all the iron in the ore is present as Fe3O5. We can then calculate the percentage based on the molar mass of Fe3O5.
1. Calculate the molar mass of Fe3O5.
Molar mass of Fe3O5 = (3 * atomic mass of Fe) + (5 * atomic mass of O)
= (3 * 55.845 g/mol) + (5 * 16.00 g/mol)
= 221.52 g/mol
2. Calculate the mass of Fe3O5 corresponding to the moles of Fe3+ ions produced.
Mass of Fe3O5 = Moles of Fe3+ * Molar mass of Fe3O5
= 0.001107942 moles * 221.52 g/mol
= 0.245481 g
3. Calculate the percentage by mass of Fe3O5 in the ore.
Percentage by mass of Fe3O5 = (Mass of Fe3O5 / Mass of the ore sample) * 100
= (0.245481 g / 1.362 g) * 100
= 18.01%
Therefore, the percentage by mass of Fe3O5 in the ore is approximately 18.01%.