Question #0d581

1 Answer
Jun 22, 2017

"FeO " : " Fe"_2"O"_3 = 4:3FeO : Fe2O3=4:3

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 xx to be the number of moles of iron that reacts to form iron(II) oxide and yy the number of moles of iron that reacts to form iron(III) oxide, you can say that you have

x + y = 1" "color(darkorange)((1))x+y=1 (1)

Notice that the first reaction consumes iron and oxygen gas in a 2:12:1 mole ratio. This implies that the number of moles of oxygen gas consumed to form iron(II) oxide is equal to

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 4:3 mole ratio. This implies that the number of moles of oxygen gas consumed to form iron(III) oxide is equal to

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 color(darkorange)((1)) to write

x = 1-y

Plug this into equation color(darkorange)((2)) to get

(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