Question

When limestone, which is principally $CaC{O_3}$ , is heated, carbon dioxide and quicklime are produced by the reaction?
$CaC{O_3}(s) \to CaO(s) + C{O_2}(g)$
If $10.6 \cdot g\;$ of $C{O_2}$ was produced from the thermal decomposition of $44.14 \cdot g\;$ of $CaC{O_3}$ , what is the percent yield of the reaction?

Answer 1

For calculating the percent yield, we need to find out the theoretical and actual yields first. The theoretical yield is the mass of carbon dioxide produced when complete degradation of $44.14 \cdot g\;$ of $CaC{O_3}$ occurs with $100\%$ yield. For calculating percent yield, the actual yield is divided by theoretical yield and then multiplied by $100$ .

Complete answer:
Calculation of theoretical yield –
The balanced chemical equation indicates that when one mole of calcium carbonate undergoes thermal decomposition at $100\%$ yield, one mole of carbon dioxide is produced.
Using the molar mass of calcium carbonate i.e. $100.09g/mol$ to convert the mass of the sample to moles.
$44.14g \times \dfrac{{1 \cdot mole \cdot CaC{O_3}}}{{100.09g}} = 0.4410 \cdot moles \cdot CaC{O_3}$
Theoretically, the $0.4410 \cdot mole$ of carbon dioxide, the equivalent of – using the molar mass of carbon dioxide here –
$0.4410 \cdot moles \cdot C{O_2} \times \dfrac{{44.01{\text{ }}g}}{{1 \cdot mole \cdot C{O_2}}} = 19.41{\text{ }}g$
So at $100\%$ yield, thermal decomposition of $44.14 \cdot g\;$ of calcium carbonate produces $19.41 \cdot g$ of carbon dioxide.
Now, as given in question the actual yield is $10.6 \cdot g\;$ and theoretical is found to be $19.41 \cdot g$ , so the percent yield can be calculated as –
$\% yield = \dfrac{{10.6g}}{{19.41g}} \times 100\% = 54.6\%$
Hence, the percent yield is found to be $54.6\%$ .

Note:
The decomposition reaction is a chemical reaction in which a single compound is decomposed into two or more new elements or compounds. Calcium carbonate or limestone decomposes into calcium oxide and carbon dioxide when heated, and reacts to produce quicklime gas and carbon dioxide.