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#
"Iron(III) oxide reacts with carbon monoxide according to the
equation:
A reaction mixture initially contains 22.55 g
what mass (in g) of the excess reactant remains?"
The textbook says the answer is 2.91 g of
22.55g
14.68g
.2171mol of
.5277mol of
Therefore,
So how is the answer 2.91g of
1 Answer
It's not.
#"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 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
That is, if
#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
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...
