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CHEM5 hard Transition Metals Question help ?????? watch

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    In the presence of dilute sulphuric acid, manganate( VII) ions oxidise ethanedioate ions to carbon dioxide.

    A 2.025 g sample of hydrated iron(II) ethanedioate, FeC2O4.xH2O, was dissolved in dilute sulphuric acid and made up to 250 cm3 in a volumetric flask. A 25.0 cm3 sample of the solution was titrated at 70°C with 0.0200 M potassium manganate( VII) solution. The titration required a volume of 22.5 cm3 to produce the first permanent pink colour. Calculate the value of x.
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    (Original post by Khanster124)
    In the presence of dilute sulphuric acid, manganate( VII) ions oxidise ethanedioate ions to carbon dioxide.

    A 2.025 g sample of hydrated iron(II) ethanedioate, FeC2O4.xH2O, was dissolved in dilute sulphuric acid and made up to 250 cm3 in a volumetric flask. A 25.0 cm3 sample of the solution was titrated at 70°C with 0.0200 M potassium manganate( VII) solution. The titration required a volume of 22.5 cm3 to produce the first permanent pink colour. Calculate the value of x.
    In this kinds of questions, you're most likely gonna need to calculate the moles of a substance to get the moles of the other substance.

    We can calculate the moles of Potassium Manganate using Moles = Volume x Concentration.

    We're told that ethanedioate ions react with manganate ions. Figure out the full equation so that you know the mole ratio between ethanedioate ions and manganate ions.
    Spoiler:
    Show
    C2O4 ----> 2CO2 + 2e-
    MnO4 + 8H+ + 5e- ----> Mn2+ + 4H2O
    Now that we have the mole ratio, you can calculate the moles of ethanedioate ions based on the moles of manganate ions.

    Note this is talking about a 25cm3 sample of iron ethanedioate. Therefore, we need to calculate the number of moles in the original sample, i.e. 250cm3
    Think, if 25cm3 = x moles, then 250cm3 = ____ moles

    So now that we have the original moles of ethanedioate ions, this is also equal to the moles of iron ethanedioate as FeC2O4xH2O ---> Fe2+ + C2O42- + xH2O

    To calculate x, we have to use the mass given. Use the equation Mass = Mr x Mole.
    You know the mass, and you know the moles, so use the two to calculate the Mr of this compound.

    Finally, deduct the Mr of the known parts of the compound. What I mean by this, is that we know the Mr of 'Fe' and 'C'2O4 so deduct the Mr of Fe and C2O4
    and what you will be left with is 18x (18 is the Mr of H2O)
    Divide by 18 to be left with x.
    Spoiler:
    Show
    Moles of Manganate = 0.00045 mol
    2:5 ratio ∴ Moles of ethanedioate = 0.001125 mol
    Moles in 250cm3 = 0.01125 mol
    Total Mr = 180.0
    ∴ 55.9 + 2(12) + 4(16) + 18x = 180
    ∴ 18x = 36.1
    ∴ x = 2

    My maths/science could be off so if someone could verify that would be appreciated.
 
 
 
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