Question

Question #878d6

Answer 1

Molar mass of
`"Fe"_2"O"_3=(2xx56+3xx16)g"/mol"=160g"/mol"`

So `160g" "Fe_2O_3` contains `112g" " Fe`

Hence `160"tonne "Fe_2O_3` contains `112"tonne " Fe`

`"Fe"_2"O"_3(s) + 3"CO"(g) -> 2"Fe"(l) + 3"CO"_2(g)`

As per the given balanced equation 160 g of `Fe_2O_3` produces 112 g `Fe` on reduction.

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