Question #9701f

1 Answer
Oct 22, 2017

Answer:

#72.1%#

Explanation:

The first thing that you need to do here is to figure out how much carbon dioxide should be produced when the reaction consumes #"75.0 g"# of carbon monoxide.

This amount will represent the reaction's theoretical yield, which is what you expect the reaction to produce at #100%# yield.

The balanced chemical equation

#"Fe"_ 2"O"_ (3(s)) + 3"CO"_ ((g)) -> 2"Fe"_ ((s)) + 3"CO"_ (2(g))#

tells you that when the reaction consumes #3# moles of carbon monoxide, it should produce #3# moles of carbon dioxide, provided, of course, that you have at least #1# mole of iron(III) oxide available.

Use the molar masses of carbon monoxide and of carbon dioxide to convert this #3:3# mole ratio, which is equivalent to a #1;1# mole ratio, to a gram ratio.

For carbon monoxide, you will have

#3 color(red)(cancel(color(black)("moles CO"))) * "28.01 g"/(1color(red)(cancel(color(black)("mole CO")))) = "84.03 g"#

Similarly, for carbon dioxide, you will have

#3 color(red)(cancel(color(black)("moles CO"_2))) * "44.01 g"/(1color(red)(cancel(color(black)("mole CO"_2)))) = "132.03 g"#

This means that when the reaction consumes #"84.03 g"# of carbon monoxide, it should theoretically produce #"132.03 g"# of carbon dioxide.

Use this gram ratio to calculate the theoretical yield for your reaction.

#75.0 color(red)(cancel(color(black)("g CO"))) * overbrace("132.03 g CO"_2/(84.03color(red)(cancel(color(black)("g CO")))))^(color(blue)("equivalent to the 3:3 mole ratio")) = "117.84 g CO"_2#

Now, you know that this reaction actually produced #"85.0 g"# of carbon dioxide. This represents the reaction's actual yield.

In order to find the percent yield, you need to divide the actual yield by the theoretical yield and multiply the ratio by #100%#.

You will have

#"% yield" = (85.0 color(red)(cancel(color(black)("g"))))/(117.84color(red)(cancel(color(black)("g")))) * 100% = color(darkgreen)(ul(color(black)(72.1%)))#

The answer is rounded to three sig figs.

You can thus say that your reaction produces #"72.1 g"# of carbon dioxide for every #"100 g"# of carbon dioxide that it could theretically produce.