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# Redox titration help...

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1. a) Combine the two half-equations (given in the introduction) to give the overall redox equation for the reaction that has taken place during the titration.

MnO4- + 8H+ + 5Fe3+ = 5Fe2+ + Mn2+ + 4H2O

b) Use your overall equation to determine the ratio of moles of manganate(VII) ions that react with iron(II) ions.

1: 5

c) Use the average titre to calculate the moles of manganate(VII) ions which have been used in the titration.
Average titre was 18.20 cm3
concentration of manganite ions : 0.0200g
number of moles = 0.0200 * 18.20/1000 = 3.64 x 10-4

d) Calculate the amount, in moles, of iron(II) ions in the 25 cm3 sample of iron(II) sulfate.

3.64 x 10-4 * 5 = 1.82 x 10-3 mol

e) Calculate the amount, in moles, of iron(II) ions in the 100 cm3 graduated flask at the start of the experiment.
1.82 x 10-3 x (100/25)= 7.28 x 10-3

f) Calculate the mass of Fe in the original five iron tablets and hence the mass of Fe in one iron tablet.
n x mr = m
7.28 x 10-3 * 55.8 = 0.41g

0.41g / 5 = 0.08g

CAN SOMEONE PLEASE CHECK THESE ANSWERS?? THANKYOU XX
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3. Without doing the rest of the question (about to go to a meeting...) your first equation looks wrong. the Fe2+ should be on the left hand side of the equation. it should read:

MnO4- + 8H+ + 5Fe2+ = 5Fe3+ + Mn2+ + 4H2O

Hope that helps!
4. (Original post by ajsullivan)
Without doing the rest of the question (about to go to a meeting...) your first equation looks wrong. the Fe2+ should be on the left hand side of the equation. it should read:

MnO4- + 8H+ + 5Fe2+ = 5Fe3+ + Mn2+ + 4H2O

Hope that helps!
LOL thanks

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Updated: April 11, 2016
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