Question #0d581

1 Answer
Jun 22, 2017

Answer:

#"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#