Question

When limestone rock, which is principally calcium carbonate, is heated, a reaction occurs. If 11.7 g of carbon dioxide were produced in the lab from the decomposition of 30.7g of calcium carbonate, what is the percent yield for the reaction?

Answer 1

Answer: 86.7 %

Explanation:

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}$

a) moles of $CaCO_3$

$\text{Number of moles}=\frac{30.7g}{100g/mol}=0.307moles$

$CaCO_3\rightarrow CaO+CO_2$

According to stoichiometry :

1 moles of $CaCO_3$ give = 1 mole of $CO_2$

Thus 0.307 moles $CaCO_3$ give =$\frac{1}{1}\times 0.307=0.307moles$ of $CO_2$

Mass of $CO_2=moles\times {\text {Molar mass}}=0.307moles\times 44g/mol=13.5g$

$\%\text{ yield }=\frac{\text{Actual yield}}{\text{Theoretical yield }}\times 100=\frac{11.7g}{13.5g}\times 100=86.7\%$

Therefore, the percent yield for the reaction is, 86.7%