Question

Combining 0.265 mol of `Fe_2O_3` with excess carbon produced 14.7 g of Fe. `Fe_2O_3 + 3C -> 2Fe +3CO.` What is the actual yield of iron in moles?

Answer 1

Approx. `50%`

Explanation:

You have the chemical equation, which shows that each mole of ferric oxide should yield 2 moles of iron metal:

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

`"Moles of iron"` `=` `(14.7*g)/(55.8*g*mol^-1)` `=` `0.263*mol`

Because stoichiometric reaction would have yielded `2xx0.265*mol` of iron metal, the yield of the reaction is just under 50%.