Question

Chlorine forms from the reaction of hydrochloric acid with manganese(IV) oxide. Calculate the theoretical yield and the percent of chlorine if 86.0g of MnO2 and 50.0g of HCl react. The actual yield of Cl2 is 20.0g.

Answer 1

`\large \boxed {\text{24.3 g Cl}_(2); 82.3 \%}`

Step-by-step 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

`\text{Moles of MnO}_(2) = \text{86.0 g MnO}_(2) * \frac{\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} * \frac{\text{1mol HCl }}{\text{36.46 g HCl }} = \text{1.371 mol HCl}`

3. Calculate the moles of Cl₂ formed from each reactant

From MnO₂:

`\text{Moles of Cl$_(2)$} = \text{0.9892 mol MnO}_(2) * \frac{\text{1 mol Cl$_(2)$}}{\text{1 mol MnO}_(2)} = \text{0.9892 mol Cl}_(2)`

From HCl:

`\text{Moles of Cl$_(2)$} = \text{1.371 mol HCl} * \frac{\text{1 mol Cl$_(2)$}}{\text{4 mol HCl}} = \text{0.3428 mol Cl}_(2)`

4. Identify the limiting reactant

The limiting reactant is HCl, because it forms fewer moles of Cl₂.

5. Calculate the theoretical yield of Cl₂

`\text{ Mass of Cl$_(2)$} = \text{0.3428 mol Cl$_(2)$} * \frac{\text{70.91 g Cl$_(2)$}}{\text{1 mol Cl$_(2)$}} = \textbf{24.3 g Cl}_\mathbf{{2}}\\\\\text{The theoretical yield is $\large \boxed{\textbf{24.3 g Cl}_\mathbf{{2}}}$}`

6. Calculate the percentage yield of Cl₂

`\text{Percentage yield} = \frac{\text{Actual yield}}{\text{Theoretical yield}} * 100 \, \%= \frac{\text{20.0 g}}{\text{24.3 g}} * 100 \, \% = 82.3 \, \%\\\text{The percentage yield is $\large \boxed{\mathbf{82.3 \, \%}}$}`