Combining Equations AS CHEMISTRY HELP

Sooo I am working through my sheet of combining equations and I need help with the final step of combining equations:
The question is:

1) MnO4()- + 4Mn2+ + 8H+ → 5Mn(3+) + 4H2O
2) 2Mn(3+) + C2O4(2-) → 2Mn(2+) + 2CO(2)

So the first thing i did was to identify the products in the first reaction which are reactants in the second reaction in this case 5Mn(3+) & 2Mn(3+)

To make the euqation balanced I multiplied up the 2) equation so that 2Mn(3+) became 10Mn(3+) Multiplying by 5

I then multiplied the first equation by 2 so that 5Mn(2+) became 10Mn(2+)

I then wrote out both equations with all reactants from both equations together and all the products from both equations together after all the crossing anything that appears on both sides I was left with what i thought to be the right answer of:


10MnO4(-) + 18Mn(2+) + 16H+ ---> 8H20 + 5C(2)O4(2-) + 10CO(2)

But the right answer is:

2MnO4- + 5C2O42- + 16H+ → 2Mn2+ + 8H2O + 10CO2

Can anyone please explain to me what im doing wrong or what im missing.
Any help is very much appreciated,
Thank You

1) MnO4()- + 4Mn2+ + 8H+ → 5Mn(3+) + 4H2O
2) 2Mn(3+) + C2O4(2-) → 2Mn(2+) + 2CO(2)

So the first thing i did was to identify the products in the first reaction which are reactants in the second reaction in this case 5Mn(3+) & 2Mn(3+)

To make the euqation balanced I multiplied up the 2) equation so that 2Mn(3+) became 10Mn(3+) Multiplying by 5

I then multiplied the first equation by 2 so that 5Mn(2+) became 10Mn(2+)


10MnO4(-) + 18Mn(2+) + 16H+ ---> 8H20 + 5C(2)O4(2-) + 10CO(2)

You're along the right lines, thinking about it in two stages and considering what are reactants and what are products.

The method you use of multiplying each equation is also a good method to go by.

Perhaps you can see your answer cannot be right, from the fact that Carbon doesn't appear on the left, but is present on the right.

So, from your method you know that:

2MnO₄^- + 8Mn²⁺ + 16H⁺ → 10Mn³⁺ + 8H₂O

and also that:

10Mn³⁺ + 5C₂O₄^2- → 10Mn²⁺ + 10CO₂

so all the reactants together are

2MnO₄^- + 8Mn²⁺ + 16H⁺ + 10Mn³⁺ + 5C₂O₄^2-

And all the products together are

10Mn³⁺ + 8H₂O + 10Mn²⁺ + 10CO₂

The Mn³⁺ all cancel and so do 8 of the Mn²⁺

So overall

2MnO₄^- + 16H⁺ + 5C₂O₄^2- → 8H₂O + 2Mn²⁺ + 10CO₂

I think your mistake probably happened when grouping together all the reactants and all the products, so Oxalate and Mn²⁺ from equation 2) have ended up on the wrong side of the final equation. This might be because of mixing up Mn²⁺ and Mn³⁺

Hope that helps