#"13.1 g CaO"# react with #"38.9 g CO"_2"#. What is the percent yield if #"16.7 g CaCO"_3"# are produced?

Equation in balance.

The following steps will be involved in the process:

We can be certain that all of the mole ratios between the compounds are 1:1 because there are no coefficients.

We must divide the given masses by the molar masses of each reactant to find the moles of each one.

To find the percent yield, we first need to calculate the theoretical yield of calcium carbonate (CaCO3) using the given amounts of calcium oxide (CaO) and carbon dioxide (CO2). Then, we compare the actual yield (16.7 g) with the theoretical yield to find the percent yield.

1. Calculate the molar masses:

   • CaO: 40.08 g/mol (Ca: 40.08 g/mol, O: 16.00 g/mol)
   • CO2: 44.01 g/mol (C: 12.01 g/mol, O: 16.00 g/mol)

2. Determine the limiting reactant by comparing the moles of CaO and CO2.

   • Moles of CaO: $\frac{13.1 \text{ g}}{40.08 \text{ g/mol}} = 0.327 \text{ mol}$
   • Moles of CO2: $\frac{38.9 \text{ g}}{44.01 \text{ g/mol}} = 0.884 \text{ mol}$

3. From the stoichiometry of the reaction $\text{CaO} + \text{CO}_2 \rightarrow \text{CaCO}_3$, we see that 1 mole of CaO reacts with 1 mole of CO2 to produce 1 mole of CaCO3.

4. Since the moles of CaO and CO2 are equal in this case, CaO is the limiting reactant.

5. Calculate the theoretical yield of CaCO3 using the moles of CaO:

   • Theoretical yield = moles of limiting reactant × molar mass of CaCO3
   • Theoretical yield = $0.327 \text{ mol} \times 100.09 \text{ g/mol} = 32.73 \text{ g}$

6. Calculate the percent yield:

   • Percent yield = $\frac{\text{Actual yield}}{\text{Theoretical yield}} \times 100\%$
   • Percent yield = $\frac{16.7 \text{ g}}{32.73 \text{ g}} \times 100\% \approx 51\%$

Therefore, the percent yield of calcium carbonate is approximately 51%.