A 1.00g of sample of an iron ore containing Fe2O3 was dissolved in an acid and all the iron converted to Fe^2+. The solution was to rated with 90.40 cm^3 of 0.2M to give Me ^3+ and MnO4^- among the products.
a) Balance equation using the iron -electron method.
b) Calculate the mass of Fe2O3 in the sample
c) What percentage of ore was Fe2O3
a) To balance the equation using the iron-electron method, follow these steps:
Step 1: Identify the oxidation states of each element in the reaction. In this case, we have:
Fe2O3: Iron is in an oxidation state of +3.
Fe^2+: Iron is in an oxidation state of +2.
Me^3+: Me is in an oxidation state of +3.
MnO4^-: Manganese is in an oxidation state of +7 and oxygen is in an oxidation state of -2.
Step 2: Write down the unbalanced equation:
Fe2O3 + Me^3+ + MnO4^- → Fe^2+ + Me^2+ + Mn^2+ + H2O
Step 3: Balance the atoms other than oxygen and hydrogen:
2Fe2O3 + 3Me^3+ + MnO4^- → 2Fe^2+ + 3Me^2+ + Mn^2+ + H2O
Step 4: Balance the oxygen atoms by adding H2O molecules:
2Fe2O3 + 3Me^3+ + MnO4^- → 2Fe^2+ + 3Me^2+ + Mn^2+ + 4H2O
Step 5: Balance the hydrogen atoms by adding H+ ions:
2Fe2O3 + 3Me^3+ + MnO4^- + 8H+ → 2Fe^2+ + 3Me^2+ + Mn^2+ + 4H2O
b) To calculate the mass of Fe2O3 in the sample, we need to use stoichiometry. Here's how:
Step 1: Determine the number of moles of Fe^2+ reacted.
The reaction stoichiometry shows that each mole of Fe2O3 reacts with 2 moles of Fe^2+. Since we know that 90.40 cm^3 (0.09040 L) of 0.2 M solution of Fe^2+ was used, we can calculate the number of moles of Fe^2+:
0.09040 L * 0.2 mol/L = 0.01808 mol Fe^2+
Step 2: Determine the number of moles of Fe2O3 reacted.
Since each mole of Fe2O3 reacts with 2 moles of Fe^2+, the number of moles of Fe2O3 reacted is half the number of moles of Fe^2+:
0.01808 mol Fe^2+ * (1 mol Fe2O3 / 2 mol Fe^2+) = 0.00904 mol Fe2O3
Step 3: Calculate the mass of Fe2O3 in grams using the molar mass of Fe2O3, which is 159.69 g/mol:
Mass of Fe2O3 = 0.00904 mol * 159.69 g/mol = 1.444 g
Therefore, the mass of Fe2O3 in the sample is 1.444 grams.
c) To determine the percentage of the ore that was Fe2O3:
Step 1: Calculate the mass of the sample used.
We were given that the sample has a mass of 1.00 g.
Step 2: Calculate the percentage of Fe2O3 in the sample using the formula:
Percentage of Fe2O3 = (Mass of Fe2O3 / Mass of sample) * 100
Percentage of Fe2O3 = (1.444 g / 1.00 g) * 100 = 144.4%
Therefore, the percentage of ore that was Fe2O3 is 144.4%.
a) To balance the equation using the iron-electron method, we need to consider the oxidation states of iron.
The iron in Fe2O3 is in the +3 oxidation state and is reduced to the +2 oxidation state in Fe^2+. The Mn in MnO4^- is in the +7 oxidation state and is reduced to the +2 oxidation state. The Me in Me^3+ is in the +1 oxidation state and is oxidized to the +2 oxidation state.
The balanced equation is as follows:
10Fe2O3 + 8Me^3+ + MnO4^- → 10Fe^2+ + 8Me^2+ + Mn^2+ + 12H2O
b) To calculate the mass of Fe2O3 in the sample, we need to use the stoichiometry of the balanced equation.
From the equation, we can see that 10 moles of Fe2O3 react with 10 moles of Fe^2+. The molar mass of Fe2O3 is:
Fe2O3 = (2 x atomic mass of Fe) + (3 x atomic mass of O)
= (2 x 55.85 g/mol) + (3 x 16.00 g/mol)
= 111.70 g/mol + 48.00 g/mol
= 159.70 g/mol
Since 10 moles of Fe2O3 react with 10 moles of Fe^2+, we can use the molar mass of Fe2O3 to calculate the mass:
Mass of Fe2O3 = (10 moles) x (159.70 g/mol)
= 1597 g
Therefore, the mass of Fe2O3 in the sample is 1597 g.
c) To find the percentage of Fe2O3 in the ore, we need to calculate the mass percentage.
Mass percentage of Fe2O3 = (Mass of Fe2O3 / Total mass of ore) x 100
Since we have 1.00 g of ore, the total mass of the ore is 1.00 g.
Mass percentage of Fe2O3 = (1597 g / 1.00 g) x 100
= 159700%
Therefore, the percentage of Fe2O3 in the ore is 159700%.