Answer:
[tex]\large \boxed {\text{24.3 g Cl}_{2}; 82.3 \%}[/tex]
Explanation:
We are given the masses of two reactants and asked to determine the mass of the product.
This looks like a limiting reactant problem.
1. Assemble the information
We will need a balanced equation with masses and molar masses, so let’s gather all the information in one place.
MM: 86.94 36.46 70.91
MnO₂ + 4HCl ⟶ MnCl₂ + Cl₂ + 2H₂O
Mass/g: 86.0 50.0 20.00
2. Calculate the moles of each reactant
[tex]\text{Moles of MnO}_{2} = \text{86.0 g MnO}_{2} \times \dfrac{\text{1 mol MnO}_{2}}{\text{86.94 g MnO}_{2}} = \text{0.9892 mol MnO}_{2}\\\\\text{Moles of HCl} = \text{50.0 g HCl} \times \dfrac{\text{1mol HCl }}{\text{36.46 g HCl }} = \text{1.371 mol HCl}[/tex]
3. Calculate the moles of Cl₂ formed from each reactant
From MnO₂:
[tex]\text{Moles of Cl$_{2}$} = \text{0.9892 mol MnO}_{2} \times \dfrac{\text{1 mol Cl$_{2}$}}{\text{1 mol MnO}_{2}} = \text{0.9892 mol Cl}_{2}[/tex]
From HCl:
[tex]\text{Moles of Cl$_{2}$} = \text{1.371 mol HCl} \times \dfrac{\text{1 mol Cl$_{2}$}}{\text{4 mol HCl}} = \text{0.3428 mol Cl}_{2}[/tex]
4. Identify the limiting reactant
The limiting reactant is HCl, because it forms fewer moles of Cl₂.
5. Calculate the theoretical yield of Cl₂
[tex]\text{ Mass of Cl$_{2}$} = \text{0.3428 mol Cl$_{2}$} \times \dfrac{\text{70.91 g Cl$_{2}$}}{\text{1 mol Cl$_{2}$}} = \textbf{24.3 g Cl}_\mathbf{{2}}\\\\\text{The theoretical yield is $\large \boxed{\textbf{24.3 g Cl}_\mathbf{{2}}}$}[/tex]
6. Calculate the percentage yield of Cl₂
[tex]\text{Percentage yield} = \dfrac{\text{Actual yield}}{\text{Theoretical yield}} \times 100 \, \%= \dfrac{\text{20.0 g}}{\text{24.3 g}} \times 100 \, \% = 82.3 \, \%\\\text{The percentage yield is $\large \boxed{\mathbf{82.3 \, \%}}$}[/tex]
