This is a hard one. I'll do my best to try and translate =)
A sour liquid mixture (many H+) contains Fe2+ ions and is being titrated with 0,0202M MnO4-. At the end of the titration, the amount of MnO4- that had been used up was 20cm3. How much was the mass of Fe2+ in the first mixture?
I've counted out the amount of mol for the MnO4-, and then used that against the mol/g of Fe2+, but it doesn't add up.
mols MnO4^- = 0.0202 M x 0.020 L = ??Next, write the equation for the titration rxn.
MnO4^- + Fe^+2 ==> Mn^+2 + Fe^+3
You can break it down into the half reactions and write
MnO4^- ==> Mn^+2 and
Fe^+2 ==> Fe^+3
then balance but the end result will be MnO4^- + 5Fe^+2 ==> 5Fe^+3 + Mn^+2
So mols Fe^+2= ??mols MnO4^-(from above) x (5 mols Fe^+2/1 mol MnO4^-) = xx mols Fe^+2.
g Fe^+2 = xx mols Fe^+2 x molar mass Fe.
I hope this helps. If you get stuck and need further assistance, please post your work. It will help me know where you are having a problem.
To solve this problem, you first need to calculate the number of moles of MnO4- that were used in the titration.
Given:
Molarity of MnO4- solution = 0.0202 M
Volume of MnO4- solution used = 20 cm3 = 0.020 L
To calculate the moles, use the formula:
moles = concentration x volume
moles MnO4- = 0.0202 M x 0.020 L = 0.000404 moles
Next, you need to write the balanced equation for the titration reaction:
MnO4- + 5Fe2+ -> 5Fe3+ + Mn2+
From the equation, you can see that 1 mole of MnO4- reacts with 5 moles of Fe2+. So, you can use the mole ratio to find the number of moles of Fe2+.
moles Fe2+ = moles MnO4- x (5 moles Fe2+ / 1 mole MnO4-)
moles Fe2+ = 0.000404 moles x (5 moles Fe2+ / 1 mole MnO4-)
moles Fe2+ = 0.00202 moles
Now that you have the number of moles of Fe2+, you can calculate the mass of Fe2+ using its molar mass. The molar mass of Fe is approximately 55.845 g/mol.
mass Fe2+ = moles Fe2+ x molar mass Fe
mass Fe2+ = 0.00202 moles x 55.845 g/mol
mass Fe2+ = 0.112 g
Therefore, the mass of Fe2+ in the initial mixture is approximately 0.112 grams.
I hope this explanation helps! Let me know if you have any further questions.