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The problem involves calculating the percent yield of chlorine gas (\( \text{Cl}_2 \)) in the reaction between manganese dioxide (\( \text{MnO}_2 \)) and hydrochloric acid (\( \text{HCl} \)). The balanced chemical equation is given as:

\[
\text{MnO}_2 + 4 \text{HCl} \rightarrow \text{MnCl}_2 + \text{Cl}_2 + 2 \text{H}_2\text{O}
\]

### Step 1: Determine the limiting reactant
We need to find the moles of each reactant and determine which one limits the reaction. The reactants are:

- \( 28.3 \, \text{g} \, \text{MnO}_2 \)
- \( 80.0 \, \text{g} \, \text{HCl} \)

#### Molar masses:
- Molar mass of \( \text{MnO}_2 \) = \( 54.94 + 2 \times 16 = 86.94 \, \text{g/mol} \)
- Molar mass of \( \text{HCl} \) = \( 1 + 35.45 = 36.45 \, \text{g/mol} \)

Now calculate the moles of each:

1. Moles of \( \text{MnO}_2 \):
\[
\frac{28.3 \, \text{g}}{86.94 \, \text{g/mol}} \approx 0.3256 \, \text{mol}
\]

2. Moles of \( \text{HCl} \):
\[
\frac{80.0 \, \text{g}}{36.45 \, \text{g/mol}} \approx 2.1959 \, \text{mol}
\]

According to the balanced equation, 1 mole of \( \text{MnO}_2 \) reacts with 4 moles of \( \text{HCl} \). Therefore, we check the ratio of the available moles of reactants:

\[
\frac{0.3256 \, \text{mol} \, \text{MnO}_2}{1} = 0.3256 \, \text{mol} \quad \text{and} \quad \frac{2.1959 \, \text{mol} \, \text{HCl}}{4} = 0.5489 \, \text{mol}
\]

Since \( \text{MnO}_2 \) produces fewer moles, it is the limiting reactant.

### Step 2: Theoretical yield of \( \text{Cl}_2 \)
According to the balanced equation, 1 mole of \( \text{MnO}_2 \) produces 1 mole of \( \text{Cl}_2 \). Therefore, the theoretical moles of \( \text{Cl}_2 \) are equal to the moles of \( \text{MnO}_2 \):

\[
\text{Theoretical moles of } \text{Cl}_2 = 0.3256 \, \text{mol}
\]

Now, calculate the theoretical mass of \( \text{Cl}_2 \):
- Molar mass of \( \text{Cl}_2 \) = \( 2 \times 35.45 = 70.90 \, \text{g/mol} \)

\[
\text{Theoretical mass of } \text{Cl}_2 = 0.3256 \, \text{mol} \times 70.90 \, \text{g/mol} \approx 23.08 \, \text{g}
\]

### Step 3: Calculate percent yield
The actual yield of \( \text{Cl}_2 \) is given as \( 20.0 \, \text{g} \). The percent yield is calculated as:

\[
\text{Percent yield} = \left( \frac{\text{Actual yield}}{\text{Theoretical yield}} \right) \times 100 = \left( \frac{20.0 \, \text{g}}{23.08 \, \text{g}} \right) \times 100 \approx 86.63\%
\]

### Final Answer:
The percent yield of the reaction is approximately **86.63%**.

Would you like further clarification or more details?

Here are five related questions to expand on this concept:
1. How do you determine the limiting reactant in any reaction?
2. What happens if you have more than one limiting reactant?
3. How would the percent yield change if the actual yield was lower?
4. Can you explain how side reactions can affect the percent yield?
5. How would the reaction change if an excess of HCl is used?

**Tip:** Always check the units throughout your calculation to avoid errors, especially in yield problems where mass and moles are converted.

What is the % yield of for the reaction below if 28.3 g of MnO2 react with 80.0 g of HCl and form 20.0 g of Cl2 gas? MnO2 + 4 HCl → MnCl2 + Cl2 + 2 H2O

Calculate the percent yield for the reaction where 28.3 g of MnO2 reacts with 80.0 g of HCl to produce 20.0 g of Cl2 gas. This problem involves stoichiometry, limiting reactants, and the concept of percent yield, ideal for high school students in Grades 10-12.

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