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May 1, 2016
Posted by **sparkle** on Thursday, November 4, 2010 at 5:04am.

C2O42-(aq) + MnO4-(aq) → Mn2+(aq) + CO2(g)

please teach me how to do it!

- college chemistry -
**Dr Russ**, Thursday, November 4, 2010 at 6:58amThere 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

a-2b-c=2

It is then some trivial algebra to solve these

c=1

e=4

a=8

b=5/2

d=5

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 -
**Anonymous**, Tuesday, June 7, 2011 at 6:54amtits

- college chemistry -
**JOSH**, Sunday, May 18, 2014 at 2:12pmSWAG

- college chemistry -
**Cook**, Saturday, September 27, 2014 at 7:32amSchwanz ^^