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?
If
1 Answer
Explanation:
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
#"CaCO"_ (3(s)) stackrel(color(white)(acolor(red)(Delta)aaa))(->) "CaO"_ ((s)) + "CO"_ (2(g))# #uarr#
The balanced chemical equation tells you that when
Use the molar mass of calcium carbonate to convert the mass of the sample to moles.
#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
#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
However, you know that in this case, the reaction produced
#"% 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 answer is rounded to three sig figs, the number of sig figs you have for the mass of carbon dioxide.