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:

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:

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:

What is the charge in coulomb of Fe^(3+) ion

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 maintain electrons neutrality .