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))#
#4"Fe"_ ((s)) + 3"O"_ (2(g)) -> 2"Fe"_ 2"O"_ (3(s))#
Now, if you take
#x + y = 1" "color(darkorange)((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#