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Redox reaction

Posted: Sun Mar 31, 2019 10:15 am
by abo
Can somebody out here help me with this redox reaction? it says that KMnO4 reacts with K2C2O4 in acidic conditions and the products are MnO2 and CO2?

this is how I did it:

[KMnO4 + (4H+) + (3e-) -----> MnO2 + (K+) + 2H2O] *2

[K2C2O4 ------> 2CO2 + (2K+) + (2e-) ] *3
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2KMnO4 + 3K2C2O4 + (8H+) ------> 2MnO2 + 6CO2 + (8K+) + 4H2O

Posted: Sun Mar 31, 2019 11:52 pm
by ChenBeier
From the writing and calculation its ok. But in practise in acidic conditions Permanganate will be reduced to Mn2+. Manganese dioxide you get at neutral or alkaline conditions.

Posted: Mon Apr 01, 2019 12:32 am
by abo
Thank you very much for your answer. As I said the products were given in the question so I had to follow the instructions.
Anyways I have got a balanced equation and based on that I tried to calculate the mass of CO2.

I would be grateful if you could have a look at my solution :).
OBS! I used the dotted line so that every value comes under the respective chemical formula.

......2KMnO4 + 3K2C2O4 + (8H+) ------> 2MnO2 + 6CO2 + (8K+) + 4H2O
......2 mole......3 mole ............................................6 mole

.......2*158 g.....3*166 g...........................................6*44 g

given: 22 g......7 g.................................................. x g

the limiting substance is K2C2O4 and therefore i will use it to do my calculation: x = (6*44*7) / 3*166) ≈ 3.71 g

Now (measured in gram) how much methane you will get if you react the amount of CO2 you have got with H2?
CO2 + 4H2 ------> CH4 + 2H2O
1 mole....................1 mole

44 g......................16 g

3,71 g...................... s g
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s= (16*3,71)/44 ≈ 1,35g

Posted: Mon Apr 01, 2019 12:56 am
by ChenBeier
Sounds good for me.

Posted: Mon Apr 01, 2019 1:24 am
by abo
thank you so much for taking the time to look at it! :)