You run an experiment according to this equation: 2HCl+CaCO_(3)-gt H_(2)O+CaCl_(2) +CO_(2) When running your lab you started with 62 .16 grams of HCI and 50.18 grams of CaCO_(3) If, after the reaction finishes, your products have a total mass of 80.10, what was the percent yield? Round your answer to the nearest 0.01 and include units! Your Answer: square square Answer units

## Step 1: Calculate the molar masses of reactants and products<br />### The molar mass of $HCl$ is $36.46 \, \text{g/mol}$, and the molar mass of $CaCO_3$ is $100.09 \, \text{g/mol}$. These values will be used to determine the limiting reactant and theoretical yield.<br /><br />## Step 2: Determine moles of each reactant<br />### Moles of $HCl = \frac{\text{mass of } HCl}{\text{molar mass of } HCl} = \frac{62.16}{36.46} = 1.705 \, \text{mol}$. <br />Moles of $CaCO_3 = \frac{\text{mass of } CaCO_3}{\text{molar mass of } CaCO_3} = \frac{50.18}{100.09} = 0.5017 \, \text{mol}$.<br /><br />## Step 3: Identify the limiting reactant<br />### From the balanced equation, $2 \, \text{mol of } HCl$ reacts with $1 \, \text{mol of } CaCO_3$. Therefore, the required moles of $HCl$ for $0.5017 \, \text{mol of } CaCO_3$ is $2 \times 0.5017 = 1.0034 \, \text{mol}$. Since $1.705 \, \text{mol of } HCl$ is available, $CaCO_3$ is the limiting reactant.<br /><br />## Step 4: Calculate the theoretical yield of products<br />### The total mass of products equals the mass of reactants due to the law of conservation of mass. Thus, the theoretical yield is the sum of the masses of $HCl$ and $CaCO_3$: <br />Theoretical yield = $62.16 + 50.18 = 112.34 \, \text{g}$.<br /><br />## Step 5: Calculate the percent yield<br />### Percent yield is given by $\text{Percent Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100$. Substituting the values: <br />$\text{Percent Yield} = \frac{80.10}{112.34} \times 100 = 71.33\%$.