When limestone, which is principally #"CaCO"_3#, is heated, carbon dioxide and quicklime are produced by the reaction?

#"CaCO"_3(s) -> "CaO"(s) + "CO"_2(g)#

If #"10.6 g"# of #"CO"_2# was produced from the thermal decomposition of #"44.14 g"# of #"CaCO"_3#, what is the percent yield of the reaction?

Answer 1

#"% yield" = 54.6%#

The first thing that you need to do here is to figure out the theoretical yield of the reaction, which is simply the mass of carbon dioxide that would be produced by the decomposition of #"44.14 g"# of calcium carbonate at #100%# yield.

#"CaCO"_ (3(s)) stackrel(color(white)(acolor(red)(Delta)aaa))(->) "CaO"_ ((s)) + "CO"_ (2(g))# #uarr#

The balanced chemical equation tells you that when #1# mole of calcium carbonate undergoes thermal decomposition at #100%# yield, #1# mole of carbon dioxide is produced.

To convert the sample's mass to moles, use the calcium carbonate molar mass.

#44.14 color(red)(cancel(color(black)("g"))) * "1 mole CaCO"_3/(100.09 color(red)(cancel(color(black)("g")))) = "0.4410 moles CaCO"_3#

According to the aforementioned #1:1# mole ratio, the reaction should theoretically produce #0.4410# moles of carbon dioxide, the equivalent of--use the molar mass of carbon dioxide here!

#0.4410 color(red)(cancel(color(black)("moles CO"_2))) * "44.01 g"/(1color(red)(cancel(color(black)("mole CO"_2)))) = "19.41 g"#

So at #100%# yield, this reaction should produce #"19.41 g"# of carbon dioxide when it consumes #"44.14 g"# of calcium carbonate.

However, you know that in this case, the reaction produced #"10.6 g"# of carbon dioxide. To find the percent yield of the reaction, you divide the actual yield of the reaction, i.e. what it actually produces, by its theoretical yield and multiply the result by #100%#.

#"% yield" = (10.6 color(red)(cancel(color(black)("g"))))/(19.41 color(red)(cancel(color(black)("g")))) * 100% = color(darkgreen)(ul(color(black)(54.6%)))#

This essentially means that for every #"100 g"# of carbon dioxide that the reaction could theoretically produce, you will only get #"54.6 g"#.

The number of sig figs you have for the mass of carbon dioxide is the answer, rounded to three.

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Answer 2

Cooper Adams

The reaction is:

CaCO₃(s) → CaO(s) + CO₂(g)

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When limestone, which is principally #"CaCO"_3#, is heated, carbon dioxide and quicklime are produced by the reaction?

#"CaCO"_3(s) -> "CaO"(s) + "CO"_2(g)# If #"10.6 g"# of #"CO"_2# was produced from the thermal decomposition of #"44.14 g"# of #"CaCO"_3#, what is the percent yield of the reaction?

#"CaCO"_3(s) -> "CaO"(s) + "CO"_2(g)#

If #"10.6 g"# of #"CO"_2# was produced from the thermal decomposition of #"44.14 g"# of #"CaCO"_3#, what is the percent yield of the reaction?