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STATEMENT -1 : FeO is non-stoichiometri...

STATEMENT -1 : FeO is non-stoichiometric with formula `Fe_(0.95)O`.
STATEMENT -2 : Some `Fe^(2+)` ions are replaced by `Fe^(3+)` as `3Fe^(3+)` = `2Fe^(3+)` to mainatain
electrons neutrally .

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The non-stoichiometric compound Fe_0.94O is formed when some Fe^(+2) ions are replaced by Fe^(+3) ions. What is the percentage of Fe^(+3) ions in this ionic lattice ?

The non- stoichiometric compound Fe_(0.94)O is formed when x% of Fe^(2+) ions are replaced by as many 2//3 Fe^(3+) ions The value of x is:

Analyses shows that FeO has a non-stoichiometric composition with formula Fe_(0.95)O_(1.00) . Give reason.

Mole percentage of Fe^(2+) in a non-stoichiometric oxide of iroon, Fe_(0.96)O , will be: