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

Answer

B

Explanation

`1"mole "Fe_(4)O_(4)(232g)=3"mole Fe" (168g)`
Amounts of `Fe_(3)O_(4)` required for 5kg iron=`(232)/(168)xx5kg`
=6.904kg
Since, efficiency of the reaction si 85% hene, the actual required amount of `Fe_(3)O_(4)` will be
`=(100xx6.904)/(85)kg i.e.=8.12kg`