Question #5cfe2

1 Answer
Jan 29, 2016

Here's what I got.


The values given to you are actually incorrect. Here's why.

Start by writing the balanced chemical equation for this decomposition reaction.

Calcium carbonate, also known as limestone, decomposes upon heating to form calcium oxide, #"CaO"#, also known as quicklime, and carbon dioxide, #"CO"_2#

#"CaCO"_text(3(s]) stackrel(color(red)("heat")color(white)(aa))(->) "CaO"_text((s]) + "CO"_text(2(g]) uarr#

Notice that you have a #1:1# mole ratio between calcium carbonate and carbon dioxide. This should automatically tell you that the reaction will produce the same number of moles of carbon dioxide as the number of moles of calcium carbonate consumed.

Here is where percent yield comes into play. In theory, the reaction will produce one mole of carbon dioxide for every mole of calcium carbonate that undergoes decomposition.

However, this is often not what happens in real experiments. Instead of having a #100%# yield, which would correspond to the theoretical yield of the reaction, you get a smaller percent yield.

So, in essence, what the reaction actually produces is your actual yield. What the reaction should produce is your theoretical yield.

Percent yield is defined as

#color(blue)("% yield" = "actual yield"/"theoretical yield" xx 100)#

So, to find the reaction's percent yield, you need to take what is actually produced and divide it by what should have theoretically been produced.

In order to do that, you need to convert the given masses to moles by using the molar masses of the two compounds. Keep in mind that you need to convert the masses to grams first by using the conversion factor

#"1 t" = 10^3"kg" = 10^3 xx 10^3"g"#

You will have

#200 color(red)(cancel(color(black)("t"))) * (10^6color(red)(cancel(color(black)("g"))))/(1color(red)(cancel(color(black)("t")))) * "1 mole CaCO"_3/(100.09color(red)(cancel(color(black)("g")))) = 1.998 * 10^6"moles CaCO"_3#

This many moles of calcium carbonate should in theory produce #1.998 * 10^6# moles of carbon dioxide, since you have that aforementioned #1:1# mole ratio working here.

This many moles of carbon dioxide would correspond to

#1.998 * 10^6color(red)(cancel(color(black)("moles CO"_2))) * (44.01color(red)(cancel(color(black)("g"))))/(1color(red)(cancel(color(black)("mole CO"_2)))) * "1 t"/(10^6color(red)(cancel(color(black)("g")))) = "87.9 t CO"_2#

Rounded to one significant figure, the theoretical yield is equal to #"90 t"# of carbon dioxide.

So, in theory, the reaction should produce a maximum of #"90 t"# of carbon dioxide. However, you are told that the reaction produces more than that.

This is not possible. A chemical reaction cannot have a percent yield that exceeds #100%#. In this case, you would have

#"% yield" = (98 color(red)(cancel(color(black)("t"))))/(90 color(red)(cancel(color(black)("t")))) xx 100 ~~ color(red)(cancel(color(black)(109%))) -> #NOT possible

The bottom line is that percent yield problems are all about two things

  • the theoretical amount that could be produced by the reaction
  • the actual amount that is produced by the reaction