Question

Iron (III) oxide reacts with carbon monoxide to form molten iron and carbon dioxide. If .18 g of iron (III) oxide reacts with 0.11 g of carbon monoxide, how many grams of iron would be produced?

Answer 1

`Fe_2O_3(s) + 3CO(g) rarr 2Fe(l) + 3CO_2(g)`

Explanation:

`"Moles of ferric oxide "=(0.18*g)/(159.69*g*mol^-1)` `=` `1.13xx10^-3*mol" metal oxide"`.

`"Moles of carbon monoxide "=(0.11*g)/(28.0*g*mol^-1)` `=` `3.93xx10^-3*mol" CO"`.

`Fe_2O_3` is the reagent in deficiency (why?), and thus `2xx1.13xx10^-3*molxx55.85*g*mol^-1~=0.150*g` iron metal are produced.