- #1
skepticwulf
- 74
- 1
Homework Statement
When the supply of oxygen is limited, iron metal reacts with
oxygen to produce a mixture of FeO and Fe2O3. In a certain
experiment, 20.00 g iron metal was reacted with 11.20 g oxygen
gas. After the experiment, the iron was totally consumed,
and 3.24 g oxygen gas remained. Calculate the amounts of
FeO and Fe2O3 formed in this experiment.
Homework Equations
The Attempt at a Solution
3 Fe + 2 O2 ------> FeO + Fe2O3
20g 0,12mol 0,12mol
0,36mol
As each 3 mol Fe produces 1 mol FeO and Fe2O3 , just by multplying moles to their molar masses it gives:
8,64g FeO and 19,2g Fe2O3. In my solution no need for O2 in calculations.
But, solution manual disagrees :(
Firts of all it states each reaction separately (WHY?)
Then it calculates O2's mole by subtracting 3.24 from 11.2 and by dividing it to its molar mass and then:
"
Let’s assume x moles of Fe reacts to form x moles of FeO. Then 0.3581 – x, the remaining
moles of Fe, reacts to form Fe2O3. Balancing the two equations in terms of x:... " etc etc
But why?? What's wrong with my calculation? I assumed when it says "3.24 g oxygen gas remained" as O2 is surplus, excess.
Did I assume wrong?