The Fe2 (55.845 g/mol) content of a 2.271 g steel sample dissolved in 50.00 mL was determined by tiration with a standardized 0.100 M potassium permanganate (KMnO4, 158.034 g/mol) solution. The titration required 28.05 mL to reach the end point. What is the concentration of iron in the steel sample? Express your answer as grams of Fe per grams of steel (g Fe2 / g steel).
So I'm I correct in doing
mols KMnO4 = 0.1M x 0.05L = 0.005
mols Fe^2+ = 5 x 0.005 = 0.025
g Fe = 0.025mol x 55.845g/mol = 1.396125 = grams Fe in the 2.271 g sample of steel.
So convert that from g Fe in 2.271g steel to grams Fe in 1 g steel.
So to do this last part do I take the 1.396125/2.271g? To find what is in the 1g sample?
because I got 0.6147g and that is incorrect.
mols KMnO4 is M x L = 0.02805 x 0.1 =? You dissolved the sample in 50 mL, the titration with KMnO4 was 28.05 mL. Correct that step and follow through with the rest of the problem The set up looks ok for the rest of it. I ran through it and obtained approx 0.35 g Fe/g sample.
To find the concentration of iron in the steel sample, we need to use the stoichiometry of the reaction between potassium permanganate (KMnO4) and iron (Fe).
The balanced equation for the reaction is:
5Fe^2+ + MnO4^- + 8H+ -> 5Fe^3+ + Mn^2+ + 4H2O
From the equation, we can see that the ratio of Fe^2+ to KMnO4 is 5:1.
First, we calculate the number of moles of KMnO4 used.
Moles of KMnO4 = concentration of KMnO4 solution * volume of KMnO4 solution (in liters)
= 0.100 M * (28.05 mL / 1000 mL/L)
= 0.002805 moles
Since the ratio of Fe^2+ to KMnO4 is 5:1, the moles of Fe^2+ in the reaction are also 0.002805 moles.
Next, we calculate the mass of Fe in the steel sample.
Mass of Fe = moles of Fe^2+ * molar mass of Fe
= 0.002805 moles * 55.845 g/mol
= 0.1566 grams
Finally, we calculate the concentration of iron in the steel sample.
Concentration of iron = mass of Fe / mass of steel sample
= 0.1566 g / 2.271 g
= 0.069 g Fe/g steel (rounded to three decimal places)
Therefore, the concentration of iron in the steel sample is 0.069 g Fe/g steel.
I don't like the Fe2 business. Probably you mean Fe^2+ and even though that is a perfect way of writing it the Fe2 bit makes one think iron is diatomic and it isn't.
5Fe^2+ + KMnO4 ==> 6Fe^3+ + Mn^2+
(I know the equation is balanced but the redox part is and we aren't worried about the spectator ions/molecules (H^+ and H2O). You can put those in if you wish.
mols KMnO4 = M x L = ?
mols Fe^2+ = 5 x that(5 times that)
g Fe = mols Fe x atomic mass Fe = ? = grams Fe in the 2.271 g sample of steel.
So convert that from g Fe in 2.271g steel to grams Fe in 1 g steel.