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Question

0.1 mole of a metal is burnt in air it forms an oxide. The same oxide is then reduced by 4 L of 0.05 M S2O23 (acidic medium) to a +3 oxidation state of the metal. What is the oxidation state of the metal in the oxide?

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Solution

M+O2Mn+ (oxide formation)
Mn+M3+ (because of thiosulphate)
The number of equivalents need to be the same in the second redox reacion, hence:
(0.1×(n3))metal=(4×1×0.05)S2O23
n=5

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