Chlorine gas can be prepared in the laboratory by the reaction of hydrochloric acid with manganese(IV) oxide. What is the limiting reactant? \( \mathrm{HCl}_{2} \) \( \mathrm{MnO}_{2} \) What is the theoretical yield of \( \mathrm{Cl}_{2} \) ? \( 36.1 \mathrm{MnO}_{2}(\mathrm{~s}) \longrightarrow \mathrm{MnCl}_{2}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})+\mathrm{Cl}_{2}(\mathrm{~g}) \) theoretical yield: tampled to a solution containing 46.9 g HCl .

To determine the limiting reactant and the theoretical yield of chlorine gas (\( \mathrm{Cl}_2 \)) from the reaction of hydrochloric acid (\( \mathrm{HCl} \)) with manganese(IV) oxide (\( \mathrm{MnO}_2 \)), we first need to write the balanced chemical equation for the reaction. The balanced equation is: \[ \mathrm{MnO}_2(s) + 4 \mathrm{HCl}(aq) \rightarrow \mathrm{MnCl}_2(aq) + 2 \mathrm{H}_2O(l) + \mathrm{Cl}_2(g) \] From the balanced equation, we can see that: - 1 mole of \( \mathrm{MnO}_2 \) reacts with 4 moles of \( \mathrm{HCl} \) to produce 1 mole of \( \mathrm{Cl}_2 \). ### Step 1: Calculate moles of \( \mathrm{HCl} \) Given that we have 46.9 g of \( \mathrm{HCl} \), we can calculate the number of moles of \( \mathrm{HCl} \): \[ \text{Molar mass of } \mathrm{HCl} = 1.01 \, (\text{H}) + 35.45 \, (\text{Cl}) = 36.46 \, \text{g/mol} \] \[ \text{Moles of } \mathrm{HCl} = \frac{46.9 \, \text{g}}{36.46 \, \text{g/mol}} \approx 1.287 \, \text{mol} \] ### Step 2: Calculate moles of \( \mathrm{MnO}_2 \) Next, we need to determine how many moles of \( \mathrm{MnO}_2 \) are required to react with the available \( \mathrm{HCl} \). According to the balanced equation, 4 moles of \( \mathrm{HCl} \) are needed for every mole of \( \mathrm{MnO}_2 \): \[ \text{Moles of } \mathrm{MnO}_2 \text{ required} = \frac{1.287 \, \text{mol HCl}}{4} \approx 0.32175 \, \text{mol MnO}_2 \] ### Step 3: Determine the limiting reactant To determine the limiting reactant, we need to know how many moles of \( \mathrm{MnO}_2 \) we have. If we assume we have a sufficient amount of \( \mathrm{MnO}_2 \) (more than 0.32175 mol), then \( \mathrm{HCl} \) is the limiting reactant because we have calculated that we need 0.32175 mol of \( \mathrm{MnO}_2 \) to react with the available \( \mathrm{HCl} \). ### Step 4: Calculate the theoretical yield of \( \mathrm{Cl}_2 \) From the balanced equation, we see that 4 moles of \( \mathrm{HCl} \) produce 1 mole of \( \mathrm{Cl}_2 \). Therefore, the moles of \( \mathrm{Cl}_2 \) produced from the available \( \mathrm{HCl} \) can be calculated as follows: \[ \text{Moles of } \mathrm{Cl}_2 = \frac{1.287 \, \text{mol HCl}}{4} \approx 0.32175 \, \text{mol Cl}_2 \] ### Step 5: Calculate the mass of \( \mathrm{Cl}_2 \) The molar mass of \( \mathrm{Cl}_2 \) is: \[ \text{Molar mass of } \mathrm{Cl}_2 = 2 \times 35.45 \, \text{g/mol} = 70.90 \, \text{g/mol} \] Now, we can calculate the theoretical yield of \( \mathrm{Cl}_2 \): \[ \text{Mass of } \mathrm{Cl}_2 = \text{Moles of } \mathrm{Cl}_2 \times \text{Molar mass of } \mathrm{Cl}_2 \] \[ \text{Mass of } \mathrm{Cl}_2 = 0.32175 \, \text{mol} \times 70.90 \, \text{g/mol} \approx 22.83 \, \text{g} \] ### Summary - **Limiting reactant**: \( \mathrm{HCl} \) - **Theoretical yield of \(