Question

When heat is applied to 80 grams of CaCO3, it yields 39 grams of CO2. Determine

Answer 1

Answer: The percentage yield of $CO_2$ is 90.26%.

Explanation:

The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ......(1)

Given mass of $CaCO_3$ = 80 g

Molar mass of $CaCO_3$ = 100 g/mol

Plugging values in equation 1:

$\text{Moles of }CaCO_3=\frac{80g}{100g/mol}=0.8 mol$

For the given chemical equation:

$CaCO_3\rightarow CaO+CO_2$

By the stoichiometry of the reaction:

If 1 mole of $CaCO_3$ produces 1 mole of $CO_2$

So, 0.8 moles of $CaCO_3$ will produce = $\frac{1}{1}\times 0.8=0.8mol$ of $CO_2$

Molar mass of $CO_2$ = 44 g/mol

Plugging values in equation 1:

$\text{Mass of }CO_2=(0.8mol\times 44g/mol)=35.2g$

The percent yield of a reaction is calculated by using an equation:

$\% \text{yield}=\frac{\text{Actual value}}{\text{Theoretical value}}\times 100$ ......(2)

Given values:

Actual value of $CO_2$ = 35.2 g

Theoretical value of $H_2CO_3$ = 39 g

Plugging values in equation 2:

$\% \text{yield of }CO_2=\frac{35.2g}{39g}\times 100\\\\\% \text{yield of }CO_2=90.26\%$

Hence, the percentage yield of $CO_2$ is 90.26%.