Question

How is CO2 the reactant in excess in this problem?

"Iron(III) oxide reacts with carbon monoxide according to the
equation:
`Fe_2 O_3 (s) + 3 CO(g) rarr 2 Fe(s) + 3 CO_2 (g)`

A reaction mixture initially contains 22.55 g `Fe_2 O_3` and 14.78 g
`CO`. Once the reaction has occurred as completely as possible,
what mass (in g) of the excess reactant remains?"

The textbook says the answer is 2.91 g of `CO_2`. But I came up with 4.28 g of `Fe_2O_3` as the excess.

22.55g `Fe_2O_3` equals .2171 mol `Fe_2O_3`

14.68g `CO` equals .5277mol `CO`

.2171mol of `Fe_2O_3` allows for a max theoretical yield of .4343 mol of `Fe`

.5277mol of `CO` allows for a max theoretical yield of .3518mol of `Fe`

Therefore, `Fe_2O_3` is the reactant in excess. The `CO` reacts completely with 18.27g of `Fe_2O_3`, leaving 4.28g of `Fe_2O_3` in excess.

So how is the answer 2.91g of `CO_2`

Answer 1

It's not. `"CO"` or `"Fe"_2"O"_3` could be in excess... but not `"CO"_2`.

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`"Fe"_2 "O"_3 (s) + 3 "CO"(g) rarr 2 "Fe"(s) + 3 "CO"_2 (g)`

I always verify other people's work, so... if you have `"22.55 g Fe"_2"O"_3` and `"14.78 g CO"`, then...

`22.55 cancel("g Fe"_2"O"_3) xx ("1 mol")/(159.69 cancel("g Fe"_2"O"_3)) = ul("0.1412 mols Fe"_2"O"_3)`

`14.78 cancel"g CO" xx ("1 mol")/(28.01 cancel"g CO") = ul"0.5277 mols CO"`

Molar mass applies only to every `"1 mol"` of anything, NOT `"2 mols"`. As a result, we find that the mols of `"CO"` are over three times the mols of `"Fe"_2"O"_3`, and `bb"CO"` is in excess .

That is, if `"CO"` was NOT in excess, then there would be

`0.1412 cancel("mols Fe"_2"O"_3) xx ("3 mols CO")/(cancel("1 mol Fe"_2"O"_3)) = "0.4236 mols CO"`

and so there is `bb"0.1041 mols"` too much `bb"CO"`, or `bbul("2.92 g CO")`.

Therefore, we use the ferrous oxide limiting reactant to get:

`"0.1412 mols Fe"_2"O"_3 xx (3 cancel("mols CO"_2))/(cancel("1 mol Fe"_2"O"_3)) xx ("44.009 g CO"_2)/cancel("1 mol CO"_2) = ul("18.64 g CO"_2)`

as the theoretical yield...