Question

If 11.1 mol of iron(III) oxide and 22.7 mol of carbon react, how many moles of the excess reactant will remain after the reaction? 2 Fe2O3(s) + 3 C(s) → 4 Fe (s) + 3 CO2(g)

Answer 1

Consider the stoichiometric equation.....

`2Fe_2O_3(s) + 3C(s) + Delta rarr 4Fe + 3CO_2(g)uarr`. And there are approx. `72 *g` of carbon in excess.

Explanation:

Now a molar quantity of `11.1*mol` of `"ferric oxide"` REQUIRES `3/2xx11.1*mol` with respect to carbon for equivalence......i.e. `16.65*mol`. We got `22.7*mol` of carbon in the starting conditions, and thus carbon is the reagent in excess. And this makes sense, because carbon is as cheap and dirt, and we want to reduce the ferric oxide.....

And thus moles of carbon in excess `=(22.7-16.65)*mol=6.05*mol` of carbon.....