Question #9701f
1 Answer
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
This amount will represent the reaction's theoretical yield, which is what you expect the reaction to produce at
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
Use the molar masses of carbon monoxide and of carbon dioxide to convert this
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
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
In order to find the percent yield, you need to divide the actual yield by the theoretical yield and multiply the ratio by
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