Upon heating, calcium carbonate decomposes to produce calcium oxide and carbon dioxide. What is the percent yield of #CO_2# if 97.5 g #CO_2# is collected?

To calculate the percent yield of CO2, we need the theoretical yield of CO2. This can be calculated based on the stoichiometry of the reaction. Given the balanced equation:

CaCO3(s) -> CaO(s) + CO2(g)

The molar mass of CO2 is 44.01 g/mol. From the equation, 1 mole of CaCO3 produces 1 mole of CO2. Therefore, the theoretical yield of CO2 can be calculated using the molar mass of CaCO3 (100.09 g/mol) and the given mass of CO2:

$\text{Theoretical yield of CO}_2 = \frac{(\text{mass of CO}_2)}{(\text{molar mass of CO}_2/\text{molar mass of CaCO}_3)}$

$\text{Theoretical yield of CO}_2 = \frac{97.5 \, \text{g}}{44.01 \, \text{g/mol}/100.09 \, \text{g/mol}}$

$\text{Theoretical yield of CO}_2 = \frac{97.5 \, \text{g}}{0.4402}$

$\text{Theoretical yield of CO}_2 \approx 221.47 \, \text{g}$

Now, to calculate the percent yield:

$\text{Percent yield} = \frac{(\text{actual yield})}{(\text{theoretical yield})} \times 100$

$\text{Percent yield} = \frac{97.5 \, \text{g}}{221.47 \, \text{g}} \times 100$

$\text{Percent yield} \approx 44.04\%$

Therefore, the percent yield of CO2 is approximately 44.04%.