You run an experiment according to this equation: 2HCl+CaCO_(3)arrow H_(2)O+CaCl_(2) +CO_(2) When running your lab you started with 62 .16 grams of HCl 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

## Step1: Calculate the Molar Masses<br />### Determine the molar masses of $HCl$ and $CaCO_3$ using their atomic weights.<br />- $HCl$: $1.01 + 35.45 = 36.46 \, \text{g/mol}$<br />- $CaCO_3$: $40.08 + 12.01 + 3 \times 16.00 = 100.09 \, \text{g/mol}$<br /><br />## Step2: Calculate Moles of Reactants<br />### Convert the given masses of $HCl$ and $CaCO_3$ to moles.<br />- Moles of $HCl$: $\frac{62.16 \, \text{g}}{36.46 \, \text{g/mol}} = 1.705 \, \text{mol}$<br />- Moles of $CaCO_3$: $\frac{50.18 \, \text{g}}{100.09 \, \text{g/mol}} = 0.501 \, \text{mol}$<br /><br />## Step3: Determine Limiting Reactant<br />### Identify the limiting reactant by comparing the mole ratio from the balanced equation.<br />- The balanced equation shows a 2:1 ratio of $HCl$ to $CaCO_3$.<br />- Required moles of $HCl$ for 0.501 mol of $CaCO_3$: $2 \times 0.501 = 1.002 \, \text{mol}$<br />- Since we have 1.705 mol of $HCl$, $CaCO_3$ is the limiting reactant.<br /><br />## Step4: Calculate Theoretical Yield<br />### Use the limiting reactant to calculate the theoretical yield of products.<br />- Total mass of reactants: $62.16 \, \text{g} + 50.18 \, \text{g} = 112.34 \, \text{g}$<br />- Theoretical mass of products (should be equal to total mass of reactants): $112.34 \, \text{g}$<br /><br />## Step5: Calculate Percent Yield<br />### Compare the actual mass of products to the theoretical mass to find the percent yield.<br />- Actual mass of products: $80.10 \, \text{g}$<br />- Percent yield: $\left( \frac{80.10 \, \text{g}}{112.34 \, \text{g}} \right) \times 100 = 71.31\%$