Question #db883
1 Answer
Explanation:
Start by writing the balanced chemical equation that describes this reaction
#"Fe"_ 2"O"_ (3(s)) + 3"CO"_ ((g)) -> 2"Fe"_ ((s)) + 3"CO"_ (2(g))#
Now, notice that for every
This represents the reaction's theoretical yield, i.e. what you get for a reaction that has an
In your case, you know that
Use the molar mass of iron(III) oxide to convert the sample to moles
#150.0 color(red)(cancel(color(black)("g"))) * ("1 mole Fe"_2"O"_3)/(159.69color(red)(cancel(color(black)("g")))) = "0.9393 moles Fe"_2"O"_3#
Now, this much iron(III) oxide would theoretically produce
#0.9393 color(red)(cancel(color(black)("moles Fe"_2"O"_3))) * overbrace("2 moles Fe"/(1color(red)(cancel(color(black)("mole Fe"_2"O"_3)))))^(color(blue)("from the balanced chemical equation")) = "1.879 moles Fe"#
This would be equivalent to
#1.897 color(red)(cancel(color(black)("moles Fe"))) * "55.845 g"/(1color(red)(cancel(color(black)("mole Fe")))) = "104.93 g"#
So, the reaction should produce
You can thus say that the reaction has a percent yield equal to
#87.9 color(red)(cancel(color(black)("g Fe"))) * "100% yield"/(104.93 color(red)(cancel(color(black)("g Fe")))) = color(darkgreen)(ul(color(black)(83.8%)))#
The answer is rounded to three sig figs, the number of sig figs you have for the mass of iron produced by the reaction.