The amount of iron in a meteorite was determined by a redox titration using KMnO4 as the titrant. A 0.4408 g sample was dissolved in acid and the liberated Fe3+ quantitatively reduced to Fe2+. Titrating with 0.0220 M KMnO4 requires 41.38 ml to reach the end point.
Unbalance reaction: Fe2+ + MnO4- → Fe3+ + Mn2+.
Determine the % w/w Fe2O3.
Balance the equation.
mols MnO4^- = M x L = ?
Convert mols MnO4^- to mols Fe using the coefficients in the balanced equation.
g Fe = mols Fe x molar mass Fe
%Fe = (g Fe/mass sample)*100 = ?
Post your work if you get stuck.
im confused on how to balance
Mn goes from +7 in MnO4- to +2 in Mn.
Fe goes from +2 to +3.
Here is a site all about redox and balancing redox equations.
http://www.chemteam.info/Redox/Redox.html
To determine the percentage by weight of Fe2O3 in the meteorite sample, we need to follow a series of steps:
Step 1: Calculate the number of moles of KMnO4 used in the titration.
The volume of KMnO4 used is 41.38 mL, which is equal to 0.04138 L (since 1 mL = 0.001 L). The molarity (M) of KMnO4 is 0.0220 M. Therefore, we can calculate the number of moles of KMnO4 used as follows:
moles KMnO4 = Molarity × volume (in liters)
moles KMnO4 = 0.0220 mol/L × 0.04138 L = 0.00090836 mol
Step 2: Determine the moles of Fe2+ reacted with KMnO4.
From the unbalanced reaction equation, we can see that every 1 mole of KMnO4 reacts with 1 mole of Fe2+.
Therefore, the number of moles of Fe2+ is equal to the number of moles of KMnO4 used in the titration, which is 0.00090836 mol.
Step 3: Calculate the moles of Fe2O3 in the sample.
From the balanced reaction equation, we can see that 1 mole of Fe2+ reacts with 1 mole of Fe2O3.
Therefore, the number of moles of Fe2O3 is also equal to the number of moles of Fe2+ obtained in Step 2, which is 0.00090836 mol.
Step 4: Determine the mass of Fe2O3.
The molar mass of Fe2O3 is the sum of the atomic masses of iron (Fe) and oxygen (O), which can be obtained from the periodic table as follows:
Counting one Fe atom, its atomic mass is 55.845 g/mol.
Counting three O atoms, each with an atomic mass of 16.00 g/mol, the total mass for O atoms is 48.00 g/mol.
Therefore, the molar mass of Fe2O3 is 55.845 g/mol + 48.00 g/mol = 103.88 g/mol.
We can then calculate the mass of Fe2O3 using the following formula:
mass Fe2O3 = moles of Fe2O3 × molar mass of Fe2O3
mass Fe2O3 = 0.00090836 mol × 103.88 g/mol = 0.0945 g
Step 5: Calculate the percentage by weight of Fe2O3.
To determine the percentage by weight, we need to divide the mass of Fe2O3 obtained in Step 4 by the mass of the meteorite sample (0.4408 g), and multiply by 100.
% w/w Fe2O3 = (mass of Fe2O3 / mass of sample) × 100
% w/w Fe2O3 = (0.0945 g / 0.4408 g) × 100 = 21.45%
Therefore, the percentage by weight of Fe2O3 in the meteorite sample is approximately 21.45%.