[kelaklub]

The following came from a practise test and I am having trouble figuring out how they came up with the answer...
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Question
In the following reaction...

(NH4)2Cr2O7 (s) ----> Cr2O3 (s) + N2 (g) + 4H2O (g)

... the oxidation number of chromium changes from +6 to:

A) 0
B) +2
C) +3
D) +4
E) +5
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The answer is supposed to be +3.

I have no idea how they arrived at that answer or how the original oxidation number of chromium is +6.

I know all about oxidation is losing electrons or infomaly gaining oxygen, but don't know how I can apply that to this problem. Please help.

[Albert]

Ok, first of all, the answer is correct.

For what concerns the first molecule, the oxidation number is worked out in the following way: oxygen has -2 has oxidation number and there are seven atoms; hence, -14.
There are 2 cations (NH4+): hence, +2
The difference is +12, but there are 2 chromium: hence, Cr is +6.

I think you can find useful to use this scheme for understanding why your answer is correct.

[kelaklub]

Aaaah, you are absolutely correct. Thank you so much for putting it in perspective.

[Borek]

This is mostly about balancing, but all you have to do is to calculate ON - and that's described too.

http://www.chembuddy.com/?left=balancing-stoichiometry&right=oxidation-numbers-method

BTW: kelaklub, use sup and sub instead of size in formulas:

(NH₄)₂Cr₂O₇ -> Cr₂O₃ + N₂ + 4H₂O