Question #0d581
1 Answer
Explanation:
Start by writing the balanced chemical equations that describe the two reactions
2"Fe"_ ((s)) + "O"_ (2(g)) -> 2"FeO"_ ((s))2Fe(s)+O2(g)→2FeO(s)
4"Fe"_ ((s)) + 3"O"_ (2(g)) -> 2"Fe"_ 2"O"_ (3(s))4Fe(s)+3O2(g)→2Fe2O3(s)
Now, if you take
x + y = 1" "color(darkorange)((1))x+y=1 (1)
Notice that the first reaction consumes iron and oxygen gas in a
x color(red)(cancel(color(black)("moles Fe"))) * "1 mole O"_2/(2color(red)(cancel(color(black)("moles Fe")))) = x/2 "moles O"_2
Similarly, the second reaction consumes iron and oxygen gas in a
y color(red)(cancel(color(black)("moles Fe"))) * "3 moles O"_2/(4color(red)(cancel(color(black)("moles Fe")))) = (3/4y) "moles O"_2
You can thus say that you have
x/2 + 3/4y = 0.65" "color(darkorange)((2))
You now have a system of two equations with two unknowns. Use equation
x = 1-y
Plug this into equation
(1-y)/2 + 3/4y = 0.65
1/2 - (2y)/4 + (3y)/4 = 0.65
y/4 = 13/20 - 10/20
y = 3/20 * 4 = 3/5
This means that you have
x = 1 - 6/10 = 2/5
You can thus say that the first reaction will produce
2/5 color(red)(cancel(color(black)("moles Fe"))) * "2 moles FeO"/(2color(red)(cancel(color(black)("moles Fe")))) = 2/5 "moles FeO"
The second reaction will produce
3/5 color(red)(cancel(color(black)("moles Fe"))) * ("2 moles Fe"_2"O"_3)/(4color(red)(cancel(color(black)("moles Fe")))) = 3/10 "moles Fe"_2"O"_3
Therefore, the ratio between the iron(II) oxide, or ferrous oxide, and irton(III) oxide, or ferric oxide, will be
"FeO"/("Fe"_2"O"_3) = (2/5 color(red)(cancel(color(black)("moles"))))/(3/10color(red)(cancel(color(black)("moles")))) = 2/5 * 10/3 = 4/3