# Question #878d6

Jan 16, 2017

Molar mass of
$\text{Fe"_2"O"_3=(2xx56+3xx16)g"/mol"=160g"/mol}$

So $160 g \text{ } F {e}_{2} {O}_{3}$ contains $112 g \text{ } F e$

Hence $160 \text{tonne } F {e}_{2} {O}_{3}$ contains $112 \text{tonne } F e$

${\text{Fe"_2"O"_3(s) + 3"CO"(g) -> 2"Fe"(l) + 3"CO}}_{2} \left(g\right)$

As per the given balanced equation 160 g of $F {e}_{2} {O}_{3}$ produces 112 g $F e$ on reduction.

So 112tonnes of iron can be obtained by reducing 160 tonnes of iron (III) oxide.