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Topic: Manganese salt  (Read 1856 times)

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Offline Rutherford

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Manganese salt
« on: March 12, 2013, 01:52:19 PM »
Substance A is an anhydrous purple potassium salt. Substance B is the salt of one cation and one anion which in reaction with BaCl2 gives a white precipitate insoluble in dilute acid. When 7.11g of substance A reacts with B in excess in neutral environment, 9.79g of a dark brown oxide is obtained. Determine A and B.

A is certainly KMnO4. 0.045mol of it reacted to form the oxide. The anion in B is SO42-, but here I am stuck. I don't have the reaction equation. What to do now?


Also one quick question: Why is the formula Al(HSO4)3 incorrect?

Offline Borek

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Re: Manganese salt
« Reply #1 on: March 12, 2013, 03:03:54 PM »
Have you noticed 9.79 g is more than 7.11 g? It tells you something.
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Offline Rutherford

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Re: Manganese salt
« Reply #2 on: March 12, 2013, 03:43:17 PM »
The molar mass of the oxide is bigger than that of KMnO4. I can't think of anything else  ??? .

Offline Borek

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Re: Manganese salt
« Reply #3 on: March 12, 2013, 04:05:05 PM »
What is the most likely identity of the brown oxide?
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Offline Rutherford

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Re: Manganese salt
« Reply #4 on: March 12, 2013, 04:10:19 PM »
Of manganese probably, and the one that has the biggest molar mass is Mn2O7, so what now?

Offline Borek

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Re: Manganese salt
« Reply #5 on: March 12, 2013, 04:24:43 PM »
No, it is not Mn2O7, and it is quite unlikely product.

What is the most likely identity of a brown oxide that is a product of permanganate reaction?
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Offline Rutherford

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Re: Manganese salt
« Reply #6 on: March 12, 2013, 04:41:20 PM »
MnO2 is usually produces in neutral environment so it could be the oxide. It's amount is 0.1125mol, and the amount of permanganate is 0.045mol, so I don't have enough manganese in the permanganate meaning that B is MnSO4. Using the mass data, I got the following mole ratio: nKMnO4:nMnSO4:nMnO2=2:3:5. The ratio is in agreement with the redox balancing, so the reaction is:
2KMnO4+3MnSO4+2H2O :rarrow: 5MnO2+K2SO4+2H2SO4

I think that this is okay now. Thanks for the help.

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