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. 8.12 kg
B. 4.14 kg
C. 6.94 kg
D. 16.8 kg

Answer

A

Explanation

`underset(232g)(Fe_3O_4)+underset(56xx3=168g)(4CO to 3Fe)+4CO_2`
3 moles of Fe is produced from 1 mole of `Fe_3O_4`
168g of Fe will be produced from 232 g of `Fe_3O_4`
3kg of Fe will be produced from `232/168xx3000g`
=4142.8g of 4.14kg of `Fe_3O_4`