Consider the following chemical reaction : ${Fe}_3{O}_4+CO-Fe+{CO}_2.$ What mass of ${Fe}_3{O}_4$ is required to obtain $2kg$ of Fe if the process is $90\%$ efficient?

The correct option is C $3.05$
$F{e}_3{O}_4+4CO⟶3Fe+4C{O}_2$
Molecular mass of $F{e}_3{O}_4=3\times 57+64=235$ $gm$
Mass of $3$ $mole$ of $Fe=171$ $gm$
$⟹235$ $gm$ $F{e}_3{O}_4⟶171$ $gm$ $Fe$
$\left(x\right)$ $gm⟶2000$ $gm$ $Fe$
$⟹x=\frac{2000\times 235}{171}=2748.5$ $g$
As the process is $90\%$ efficient
Amount of $F{e}_3{O}_4$ required $=\frac{100}{90}\times 2748.5$ $gm=3.05$ $kg$