Question

Question #0d581

Answer 1

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

Explanation:

Start by writing the balanced chemical equations that describe the two reactions

`2"Fe"_ ((s)) + "O"_ (2(g)) -> 2"FeO"_ ((s))`

`4"Fe"_ ((s)) + 3"O"_ (2(g)) -> 2"Fe"_ 2"O"_ (3(s))`

Now, if you take `x` to be the number of moles of iron that reacts to form iron(II) oxide and `y` 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))`

Notice that the first reaction consumes iron and oxygen gas in a `2:1` 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`