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    So this is the redox equation provided
    5fe2+ + MnO4- + 8H+ --> 5Fe3+ + Mn2+ + 4H20
    So 25cm3 of a 100cm3 solution of FeSO4 is titrated against 0.02moldm-3 of 8.60cm3 of MnO4-
    So I calculated mol of mno4- and got 1.72x10-4 and multiplied this by 5 to get mOles of FeSO4
    My teacher said to calculate mass of the 100cm3 solution and using the mass of the iron tablets (1.16g) and then do mass/mass of the iron tablets but I'm really confused as to how the mass of the 100cm3 solution divided by the mass of the iron tablets tells you the percentage? Can anyone explain?
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    (Original post by AqsaMx)
    So this is the redox equation provided
    5fe2+ + MnO4- + 8H+ --> 5Fe3+ + Mn2+ + 4H20
    So 25cm3 of a 100cm3 solution of FeSO4 is pipettes again 0.02moldm-3 of 8.60cm3 of MnO4-
    So I calculated mol of mno4- and got 1.72x10-4 and multiplied this by 5 to get mOles of FeSO4
    My teacher said to calculate mass of the 100cm3 solution and using the mass of the iron tablets (1.16g) and then do mass/mass of the iron tablets but I'm really confused as to how the mass of the 100cm3 solution divided by the mass of the iron tablets tells you the percentage? Can anyone explain?
    The final answer is 16.6% btw
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    Anyone?
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    100cm3 is the intial volume.

    moles of Fe in 25cm3 = 8.6x10-4 moles
    moles Fe in 100cm3 = (8.6x10-4) x 4 = 3.44x10-3
    mass of Fe = moles x mr = (3.44x10-3) x 55.8 = 0.192 grams
    % mass = mass Fe/mass tablets x 100 = 0.192/ 1.16 x 100 = 16.54%
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    lool i realised im 3 weeks late
 
 
 

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