The formula of an oxide of iron is `Fe_(0.93)O_(1.00)`.  If the compound has hundred `O^(-2)` ions, then it contains 

Options

A. `93Fe^(+2)` ions
B. `93Fe^(+3)` ions
C. `79Fe^(+2), 14Fe^(+3)`
D. `93Fe^(+2), 14Fe^(+3)`

Answer

C

Explanation

`Fe_(0.93) O_(1.00) , Fe_(93) O_(100) , Fe^(+3) =x, Fe^(+2) = 93-x ,x xx (+3) + (93 -x) xx (+2) = 200` (or) `(2x -2)` (or) `x xx (+3) + (93 -x) xx (+2) + 100 xx (-2) =0, 3x + 186 -2x - 200 =0, x-14 =0, therefore x=14`
`Fe^(+3) = 14, Fe^(+2) = 79`