a sample iron ore is dissolved in an acidic solution in which yhe iron is converted into fe2+. the solution formed the titrated with kmno4 which oxidies fe2+ to fe3+ while kmno- ions are reduced to mn2+. 92.95 ml of 0..02M of kmno4 is required for titration to reach the equivalent point.
a) balanced eq?
b) calculate the percentage of iron in the sample.
First let me suggest you find the caps key on your keyboard and use it. There is a big difference between Co, CO, and co. kmno4 doen't mean much to me.
I'll balance the redox part; you can finish.
5Fe^2+ + KMnO4 ==> 5Fe^3+ + Mn^2+
mols KMnO4 = M x L = ?
mols Fe = mols KMnO4 x 5
grams Fe = mols Fe x atomic mass Fe
%Fe = (grams Fe/grams sample)*100 = ?
It should be obvious that you didn't give a mass of the sample so you can't calculate percent Fe in the sample.
Post your work if you get stuck.
Post your work if you get stuck.
a) The balanced equation for the reaction can be written as follows:
5 Fe2+ + 2 MnO4- + 16 H+ -> 5 Fe3+ + 2 Mn2+ + 8 H2O
b) To calculate the percentage of iron in the sample, we need to determine the number of moles of Fe2+ that reacted during the titration.
First, let's find the number of moles of KMnO4 used in the titration:
Moles of KMnO4 = concentration (M) × volume (L)
= 0.02 M × 0.09295 L
= 0.001859 moles
From the balanced equation, we can see that the stoichiometric ratio between KMnO4 and Fe2+ is 2:5. Therefore, the number of moles of Fe2+ can be calculated as follows:
Moles of Fe2+ = (5/2) × Moles of KMnO4
= (5/2) × 0.001859 moles
= 0.004648 moles
Now, we need to calculate the mass of iron in the sample:
Mass of Fe = moles of Fe × molar mass of Fe
= 0.004648 moles × 55.845 g/mol (molar mass of Fe)
= 0.2592 grams
Finally, we can calculate the percentage of iron in the sample:
Percentage of iron = (mass of Fe / mass of sample) × 100
= (0.2592 g / mass of sample) × 100
a) To balance the equation, let's first determine the oxidation states of the involved elements. In the acidic solution, Fe is oxidized from Fe2+ to Fe3+ by KMnO4, which is reduced to Mn2+.
The balanced equation is as follows:
5Fe2+ + MnO4- + 8H+ → 5Fe3+ + Mn2+ + 4H2O
b) To calculate the percentage of iron in the sample, we need to find the number of moles of KMnO4 used in the titration and relate it to the number of moles of Fe in the sample.
Given:
Volume of KMnO4 solution used (V1) = 92.95 ml = 92.95/1000 L = 0.09295 L
Molarity of KMnO4 solution (M1) = 0.02 M
Using the balanced equation, we can determine the stoichiometric ratio between KMnO4 and Fe2+ as 1:5.
Thus, the number of moles of KMnO4 used (n1) can be calculated using the formula:
n1 = M1 × V1
Substituting the given values:
n1 = 0.02 M × 0.09295 L
Next, we know that 1 mole of KMnO4 reacts with 5 moles of Fe2+, so the number of moles of Fe2+ (n2) can be calculated as:
n2 = 5 × n1
Finally, to calculate the percentage of iron in the sample, we need to relate the mass of Fe to the number of moles of Fe and the molar mass of Fe.
Molar mass of Fe = 55.845 g/mol
Mass of Fe (m) = n2 × Molar mass of Fe
To find the percentage, we use the formula:
Percentage of Fe = (m / mass of the sample) × 100
The given information does not provide the mass of the sample, so we cannot calculate the exact percentage of iron without that information.