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