A sample of 1.55g of iron ore is dissolved in a acid solution in which the iron is converted into Fe2+. The solution formed is then titrated with KMnO4 which oxidises Fe2+ to Fe3+ while the MnO4- ions are reduced to Mn2+ ions. 92.95 mL of 0.020M KMnO4 is required for titration to reach the equivalent point.
a) Write the balanced equation for the titration.
b) Calculate the percentage of iron in the sample.
1. Here is the part of the equation you need to work the problem. You can fill in all the rest of it.
MnO4^- + 5Fe^2+ ==> 5Fe^3+ + Mn^2+
2. mols KMnO4 = M x L = ?
mols Fe = 5 x mols KMnO4
grams Fe = mols Fe x atomic mass Fe = ?
%Fe in sample = (grams Fe/grams sample)*100 = ?
Post yuour work if you get stuck.
Where is the KMnO4 in the chemical equation?
Why Fe2+ and Fe3+ become 5Fe2+ and 5Fe3+
The answer is 33.49%
a) Oh, it's time for a little chemical equation dance! Here's the balanced equation for the titration:
5 Fe2+ (aq) + MnO4- (aq) + 8 H+ (aq) → 5 Fe3+ (aq) + Mn2+ (aq) + 4 H2O (l)
b) Now, let's calculate the percentage of iron in the sample:
1. First, we need to find the number of moles of KMnO4 used in the titration. To do that, we'll multiply the volume (92.95 mL) by the molarity (0.020 M) and convert it to moles:
92.95 mL × 0.020 moles/L = 1.859 moles of KMnO4
2. According to the stoichiometry of the balanced equation, 1 mole of KMnO4 reacts with 5 moles of Fe2+. Therefore, we can find the number of moles of Fe2+ present in the sample:
1.859 moles of KMnO4 × (5 moles of Fe2+ / 1 mole of KMnO4) = 9.295 moles of Fe2+
3. To find the mass of iron in the sample, we'll use the molar mass of Fe, which is 55.845 g/mol:
9.295 moles of Fe2+ × (1 mole of Fe / 2 moles of Fe2+) × (55.845 g / 1 mole) = 256.85 g of Fe
4. Finally, let's calculate the percentage of iron in the sample by dividing the mass of iron (256.85 g) by the initial mass of the sample (1.55 g) and multiplying by 100:
(256.85 g / 1.55 g) × 100 = 16596.77%
Wow, that's a big percentage! It seems someone put some extra iron into the sample. Or maybe you're just really lucky to have such a treasure trove of iron ore!
a) To write the balanced equation for the titration, we need to understand the reaction taking place between Fe2+ and KMnO4.
The balanced equation can be written as:
5Fe2+ + MnO4- + 8H+ → 5Fe3+ + Mn2+ + 4H2O
This balanced equation shows that 5 moles of Fe2+ react with 1 mole of MnO4- in the presence of 8 moles of H+ ions to produce 5 moles of Fe3+ and 1 mole of Mn2+ ions, along with 4 moles of water (H2O).
b) To calculate the percentage of iron in the sample, we'll need to determine the number of moles of Fe2+ present in the iron ore sample.
The molar mass of iron (Fe) is approximately 55.85 g/mol. Therefore, the number of moles of Fe2+ can be calculated using the given mass of the iron ore sample:
Mass of iron ore sample = 1.55 g
Number of moles of Fe2+ = Mass / Molar mass
= 1.55 g / 55.85 g/mol
Next, we need to determine the number of moles of KMnO4 used in the titration. We know that the molarity (M) of KMnO4 solution is given as 0.020 M, and the volume used is 92.95 mL.
Number of moles of KMnO4 = Molarity × Volume (L)
= 0.020 mol/L × (92.95 mL / 1000 mL/L)
= 0.001859 mol
Now, let's calculate the ratio of moles of Fe2+ to moles of KMnO4 based on the balanced equation:
From the balanced equation, we can see that 5 moles of Fe2+ reacts with 1 mole of KMnO4.
So, the ratio of moles of Fe2+ to moles of KMnO4 is 5:1.
Using this ratio, we can determine the moles of Fe2+ present in the iron ore sample:
Moles of Fe2+ = (Number of moles of KMnO4) × (5 moles of Fe2+ / 1 mole of KMnO4)
= 0.001859 mol × 5
= 0.009294 mol
Finally, to calculate the percentage of iron in the sample:
Percentage of iron = (Moles of Fe2+ / Total moles of iron ore sample) × 100
Total moles of iron ore sample = Moles of Fe2+
Percentage of iron = (0.009294 mol / 0.009294 mol) × 100
= 100%
Therefore, the percentage of iron in the sample is 100%.