Iron(III) oxide reacts with carbon monoxide according to the equation:
Fe2O3(s)+3CO(g)→2Fe(s)+3CO2(g)
A reaction mixture initially contains 22.80g Fe2O3 and 14.74g CO.
Once the reaction has occurred as completely as possible, what mass (in g) of the excess reactant is left?

Question

Iron(III) oxide reacts with carbon monoxide according to the equation: 
Fe2O3(s)+3CO(g)→2Fe(s)+3CO2(g) 
A reaction mixture initially contains 22.80g Fe2O3 and 14.74g CO.
Once the reaction has occurred as completely as possible, what mass (in g) of the excess reactant is left?

joannablackwelder · Answer

I would use stoichiometry to find the limiting reactant, first.

joannablackwelder · Answer

Then use the amount of product that is produced to calculate how much of the excess reactant is used.

anonymous · Answer

But there's two products, I've only ever done these with one product.

joannablackwelder · Answer

Use either product in your calculations, but just consistent and use whichever one you pick the whole time.

anonymous · Answer

No matter what I do, I keep getting a mass that is more than my starting mass.

joannablackwelder · Answer

It sounds like you are thinking that your limiting reactant is your excess reactant.