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college chemistry

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Balance the equation in aqueous acidic solution:
C2O42-(aq) + MnO4-(aq) → Mn2+(aq) + CO2(g)

please teach me how to do it!

  • college chemistry -

    There are two ways to balance equations
    1. by inspection
    2. by maths

    The first is usually quicker IF you can see how to do it. The second can take longer but always works.

    So to balance by inspection:

    C2O42-(aq) + MnO4-(aq) ¨ Mn2+(aq) + CO2(g)

    First make sure you are working with the correct equation, which here is

    H+ + C2O42-(aq) + MnO4-(aq) ¨ Mn2+(aq) + CO2(g) + H2O

    1. leave atoms (or ions) that are on their own to last, so here leave the Mn2+ to last

    2. include fractions if you want to but remember that the final equaion must not contain fractions

    3. Another things to notice is that in the starting equation there is an even numebr of O atoms on the LHS and an odd number on the RHS, so start by sorting this out

    H+ + C2O42-(aq) + MnO4-(aq) ¨ Mn2+(aq) + CO2(g)+ 2H2O

    4. Then balancing the H atoms

    4H+ + C2O42-(aq) + MnO4-(aq) ¨ Mn2+(aq) + CO2(g)+ 2H2O

    5. and balance the C atoms

    4H+ + C2O42-(aq) + MnO4-(aq) ¨ Mn2+(aq) + 2CO2(g)+ 2H2O

    There are now 8 O atoms on the LHS and 6 on the RHS which we can correct by putting a 4 in front of the H2O and rebalancing the H atoms

    8H+ + C2O42-(aq) + MnO4-(aq) ¨ Mn2+(aq) + 2CO2(g)+ 4H2O

    Which balances in terms of atoms but the charges do not balance as there is 5+ on the LHS and 2+ on the RHS. To balance this we need 2.5 of the C2O42- as this then gives

    (8+)+(5-)+(1-) = 2+ on the LHS which is the same as the RHS.

    So putting in the 2.5 and balancing the C atoms on the RHS.

    8H+ + 2.5C2O42-(aq) + MnO4-(aq) ¨ Mn2+(aq) + 5CO2(g)+ 4H2O

    but we are not allowed the fraction so

    16H+ + 5C2O42-(aq) + 2MnO4-(aq) ¨ 2Mn2+(aq) + 10CO2(g)+ 8H2O

    OK so how do we do this by maths?

    start by inserting symbols in front of the species, where the numbers will go

    H+ + C2O42-(aq) + MnO4-(aq) ¨ Mn2+(aq) + CO2(g) + H2O

    aH+ + bC2O42-(aq) + cMnO4-(aq) ¨ Mn2+(aq) + dCO2(g) + eH2O

    Note I have left one (Mn2+) as 1, but it does not matter which you choose.

    Next you can write equations to balance the atoms

    H. 2e=a
    C. 2b=d
    O. 4b+4c=2d+e
    Mn. c=1

    and the charge

    It is then some trivial algebra to solve these


    8H+ + 5/2C2O42-(aq) + MnO4-(aq) ¨ Mn2+(aq) + 5CO2(g) + 4H2O

    then remove the fraction by multiplying both side by 2 we get the same equation as by inspection

    16H+ + 5C2O42-(aq) + 2MnO4-(aq) ¨ 2Mn2+(aq) + 10CO2(g)+ 8H2O

  • college chemistry -


  • college chemistry -


  • college chemistry -

    Schwanz ^^

  • college chemistry -

    writing method of this question is totally wrong

  • college chemistry -

    This is not right. Do not use this answer on Mastering Chemistry.

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