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    Yes, Im in A2. But over the summer I completely forgot how to do these. I got a B, so once I'm reminded Ill be fine.

    Balance the following equation:
    Fe2+ + MnO4- + H+ ---> Fe3+ + Mn2+ + H20

    The above numbers only show the charges.

    So far I've got to:

    Fe2+ + MnO4- + 8H+ ---> Fe3+ + Mn2+ + 4H20

    Im stuck on balancing the Fe3+, 2+ and the Manganese.


    Then I have:

    Calculate the amount of moles of MnO4- in your mean titre (50cm^-3)

    I'm just puzzled here? Do I work out the amount of moles of solid MnO4- and then work out moles of solution and divide one by the other?

    Im completely puzzled and annoyed, as I'll fail A2 if i stay like this

    thanks :/
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    (Original post by fishkeeper)
    Yes, Im in A2. But over the summer I completely forgot how to do these. I got a B, so once I'm reminded Ill be fine.

    Balance the following equation:
    Fe2+ + MnO4- + H+ ---> Fe3+ + Mn2+ + H20

    The above numbers only show the charges.

    So far I've got to:

    Fe2+ + MnO4- + 8H+ ---> Fe3+ + Mn2+ + 4H20

    Im stuck on balancing the Fe3+, 2+ and the Manganese.


    Then I have:

    Calculate the amount of moles of MnO4- in your mean titre (50cm^-3)

    I'm just puzzled here? Do I work out the amount of moles of solid MnO4- and then work out moles of solution and divide one by the other?

    Im completely puzzled and annoyed, as I'll fail A2 if i stay like this

    thanks :/
    This is a redox equation. You can't balance it unless you write out the two half equations and then balance the electrons...
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    (Original post by fishkeeper)
    Yes, Im in A2. But over the summer I completely forgot how to do these. I got a B, so once I'm reminded Ill be fine.

    Balance the following equation:
    Fe2+ + MnO4- + H+ ---> Fe3+ + Mn2+ + H20

    The above numbers only show the charges.

    So far I've got to:

    Fe2+ + MnO4- + 8H+ ---> Fe3+ + Mn2+ + 4H20

    Im stuck on balancing the Fe3+, 2+ and the Manganese.


    Then I have:

    Calculate the amount of moles of MnO4- in your mean titre (50cm^-3)

    I'm just puzzled here? Do I work out the amount of moles of solid MnO4- and then work out moles of solution and divide one by the other?

    Im completely puzzled and annoyed, as I'll fail A2 if i stay like this

    thanks :/
    To make things easier, write half equations.

    5e- + 8H+ + MnO4- --> Mn2+ + 4H20
    Fe2+ --> Fe3+ +e-

    Overall Balanced:
    5Fe2+ + MnO4- + 8H+ ---> 5Fe3+ + Mn2+ + 4H20

    You can't really do much without the half equations. Remember to add electrons to one of the sides to balance the charges (they don't have to be 0, they just have to be the same).

    I assume you were given the concentration of MnO4-? You need to use that along with volume to calculate the number of moles. You can't do anything with just the volume.
 
 
 
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