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

## ${\text{CaCO"_3(s) -> "CaO"(s) + "CO}}_{2} \left(g\right)$ If $\text{10.6 g}$ of ${\text{CO}}_{2}$ was produced from the thermal decomposition of $\text{44.14 g}$ of ${\text{CaCO}}_{3}$, what is the percent yield of the reaction?

##### 1 Answer
Mar 29, 2018

"% yield" = 54.6%

#### 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 $\text{44.14 g}$ of calcium carbonate at 100% yield.

${\text{CaCO"_ (3(s)) stackrel(color(white)(acolor(red)(Delta)aaa))(->) "CaO"_ ((s)) + "CO}}_{2 \left(g\right)}$ $\uparrow$

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.

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 $1 : 1$ 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 $\text{19.41 g}$ of carbon dioxide when it consumes $\text{44.14 g}$ of calcium carbonate.

However, you know that in this case, the reaction produced $\text{10.6 g}$ 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 $\text{100 g}$ of carbon dioxide that the reaction could theoretically produce, you will only get $\text{54.6 g}$.

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