Question

Use the following equation to calculate how many moles of Fe2+(aq) were present in the conical flask.
MnO4−(aq)+8H+(aq)+5Fe2+(aq)=Mn2+(aq)+5Fe3+(aq)+8H2O(aq)
0.04 moles of MnO4- in 35.3 cm3 of FA3(0.0100 mol dm-3 potassium manganate(VII))

Answer 1

u know abt equivalence

Answer 2

no

Answer 3

okk.then any idea how u will do this

Answer 4

no clue

Answer 5

@JFraser

Answer 6

"0.04 moles of MnO4- in 35.3 cm3 of FA3(0.0100 mol dm-3 potassium manganate(VII))"

what is this supposed to mean?

Answer 7

there are 0.04 moles of MnO4 in that amount of FA3 and i said what FA3 is

Answer 8

So, what did you do with that solution?

Also, potassium manganate or potassium permanganate?

Answer 9

its potassium manganate and i used it to titrate a solution i made which was an unknown iron(II) salt and 1.00 mol dm-3 sulfuric acid.

Answer 10

Okay, so you know how many moles of potassium manganate you used. Use the coefficients to relate them to the moles of \(\sf Fe^{2+}\).

Answer 11

\(\sf \dfrac{moles~of~Fe^{2+}}{Fe^{2+} ~'s ~coefficient}=\dfrac{moles ~of~potassium~mangante}{potassium~manganate's~coefficient }\)

Answer 12

ok so is the answer 0.2?

Answer 13

according to what you posted