The Student Room Group

Chemistry A level- question

25.0 cm^3 portion of an iron(II) tablet solution required 5.00 × 10^–5 mol of manganate(VII) ions to react completely. What is the mass of iron, in grams, in the 25.0 cm^3?

Kindly explain how the answer is 0.014. Thank you in advance!
Okay if I've done this correctly, I think it should go like this:

This is a redox titration/reaction in which you have to know the half equations and combine them together. In the equation, the ratio of manganate(VII) to iron(II) is 1:5, so you need to times 0.00005 by 5 to get 0.00025 moles of iron(II)
because the volume you are trying to find mass of iron(II) is the same as the volume used in the experiment, nothing needs to be done with these values.
you then use mass = moles (0.00025) × Mr (55.8) = 0.01395 which rounds to 0.014

Quick Reply

Latest