Question #0d581

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Question #0d581

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Answer 1 · 2017-06-22T00:41:39

$FeO : {Fe}_{2} O_{3} = 4 : 3$ $"FeO " : " Fe"_2"O"_3 = 4:3$

Explanation:

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

$2 {Fe}_{(s)} + O_{2 (g)} \to 2 {FeO}_{(s)}$ $2"Fe"_ ((s)) + "O"_ (2(g)) -> 2"FeO"_ ((s))$

$4 {Fe}_{(s)} + 3 O_{2 (g)} \to 2 {Fe}_{2} O_{3 (s)}$ $4"Fe"_ ((s)) + 3"O"_ (2(g)) -> 2"Fe"_ 2"O"_ (3(s))$

Now, if you take $x$ $x$ to be the number of moles of iron that reacts to form iron(II) oxide and $y$ $y$ the number of moles of iron that reacts to form iron(III) oxide, you can say that you have

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

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

$"FeO"/("Fe"_2"O"_3) = (2/5 color(red)(cancel(color(black)("moles"))))/(3/10color(red)(cancel(color(black)("moles")))) = 2/5 * 10/3 = 4/3$

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Question #0d581

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