Redox reaction

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abo
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Redox reaction

Post 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
Last edited by abo on Tue Apr 02, 2019 3:38 am, edited 1 time in total.
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ChenBeier
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Post 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.
abo
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Post 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
----------

s= (16*3,71)/44 ≈ 1,35g
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ChenBeier
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Post by ChenBeier »

Sounds good for me.
abo
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Post by abo »

thank you so much for taking the time to look at it! :)
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