Magnetite, `Fe_(3)O_(4)`, can be converted into metallic iron by heating with carbon monoxide as represented by this equation:
`Fe_(3)O_(4)(s)+CO(g) rarr Fe(s) +CO_(2)(g)`
The kilograms of `Fe_(3)O_(4)` which must be processed in this way to obtain `5.00 kg` of iron, if the process is `85%` efficient is closest to? `[M: = Fe = 56]`

Options

A. `6.92 `kg
B. `8.12` kg
C. `20.8` kg
D. `24.4 ` kg

Answer

B

Explanation

1 mole `Fe_(3)O_(4) (232 g) = 3 " mole " Fe (168 g) `
Mmount of `Fe_(3)O_(4)` required for 5 kg iron ` = (232)/(168) xx 5 ` kg
` = 6.904 ` kg
Since, efficiency of the reaction is 85% , hence, the actual required amount of `Fe_(3)O_(4)` will
be ` = (100 xx 6.904)/85 kg , i.e. = 8.12 ` kg .