Question
Chlorine gas can be prepared in the laboratory by the reaction of hydrochloric acid with manganese(IV) oxide.
4 HCl(aq) + MnO2(s)→ MnCl2(aq) + 2H2O(l) + Cl2(g)
A sample of 36.1 g MnO, is added to a solution containing 46.9 g HCl.
What is the theoretical yield of Cl2? If the yield of the reaction is 84.5% what is the actual yield of chlorine?
4 HCl(aq) + MnO2(s)→ MnCl2(aq) + 2H2O(l) + Cl2(g)
A sample of 36.1 g MnO, is added to a solution containing 46.9 g HCl.
What is the theoretical yield of Cl2? If the yield of the reaction is 84.5% what is the actual yield of chlorine?
Answers
GPT-4o mini
To calculate the theoretical yield of chlorine gas (\(Cl_2\)), we should first determine the moles of each reactant, and then identify the limiting reagent.
### Step 1: Calculate moles of each reactant.
1. **Moles of MnO₂**:
The molar mass of \(MnO_2\) can be calculated as follows:
- Molar mass of \(Mn\) = 54.94 g/mol
- Molar mass of \(O\) = 16.00 g/mol
- Molar mass of \(MnO_2\) = 54.94 + (2 × 16.00) = 54.94 + 32.00 = 86.94 g/mol
Now, calculate the moles of \(MnO_2\):
\[
\text{Moles of } MnO_2 = \frac{36.1 \text{ g}}{86.94 \text{ g/mol}} \approx 0.414 \text{ moles}
\]
2. **Moles of HCl**:
The molar mass of \(HCl\) is approximately:
- Molar mass of \(H\) = 1.01 g/mol
- Molar mass of \(Cl\) = 35.45 g/mol
- Molar mass of \(HCl\) = 1.01 + 35.45 = 36.46 g/mol
Now, calculate the moles of \(HCl\):
\[
\text{Moles of } HCl = \frac{46.9 \text{ g}}{36.46 \text{ g/mol}} \approx 1.287 \text{ moles}
\]
### Step 2: Determine the limiting reagent.
The balanced reaction is:
\[
4 HCl(aq) + MnO_2(s) \rightarrow MnCl_2(aq) + 2H_2O(l) + Cl_2(g)
\]
From the equation, the molar ratio of \(HCl\) to \(MnO_2\) is 4:1. Therefore, we need 4 moles of \(HCl\) for every mole of \(MnO_2\).
Calculate the moles of \(HCl\) needed for 0.414 moles of \(MnO_2\):
\[
\text{Moles of } HCl \text{ needed} = 0.414 \text{ moles } MnO_2 \times 4 = 1.656 \text{ moles }
\]
Since we only have approximately 1.287 moles of \(HCl\), \(HCl\) is the limiting reagent.
### Step 3: Calculate the theoretical yield of \(Cl_2\).
From the balanced equation:
\[
4 \text{ moles of } HCl \rightarrow 2 \text{ moles of } Cl_2
\]
So, the ratio of moles of \(Cl_2\) to moles of \(HCl\) is:
\[
\text{Moles of } Cl_2 = \frac{2}{4} \times (\text{moles of } HCl) = \frac{1}{2} \times 1.287 \approx 0.6435 \text{ moles of } Cl_2
\]
### Step 4: Convert moles of \(Cl_2\) to grams.
The molar mass of \(Cl_2\) is:
\[
\text{Molar mass of } Cl_2 = 2 \times 35.45 = 70.90 \text{ g/mol}
\]
The theoretical yield of \(Cl_2\):
\[
\text{Theoretical yield of } Cl_2 = 0.6435 \text{ moles} \times 70.90 \text{ g/mol} \approx 45.7 \text{ g}
\]
### Step 5: Calculate the actual yield of chlorine.
If the yield of the reaction is 84.5%, the actual yield can be calculated as:
\[
\text{Actual yield} = \text{Theoretical yield} \times \text{percentage yield}
\]
\[
\text{Actual yield} = 45.7 \text{ g} \times 0.845 \approx 38.6 \text{ g}
\]
### Final Result:
- The theoretical yield of \(Cl_2\) is approximately **45.7 g**.
- The actual yield of \(Cl_2\) is approximately **38.6 g**.
### Step 1: Calculate moles of each reactant.
1. **Moles of MnO₂**:
The molar mass of \(MnO_2\) can be calculated as follows:
- Molar mass of \(Mn\) = 54.94 g/mol
- Molar mass of \(O\) = 16.00 g/mol
- Molar mass of \(MnO_2\) = 54.94 + (2 × 16.00) = 54.94 + 32.00 = 86.94 g/mol
Now, calculate the moles of \(MnO_2\):
\[
\text{Moles of } MnO_2 = \frac{36.1 \text{ g}}{86.94 \text{ g/mol}} \approx 0.414 \text{ moles}
\]
2. **Moles of HCl**:
The molar mass of \(HCl\) is approximately:
- Molar mass of \(H\) = 1.01 g/mol
- Molar mass of \(Cl\) = 35.45 g/mol
- Molar mass of \(HCl\) = 1.01 + 35.45 = 36.46 g/mol
Now, calculate the moles of \(HCl\):
\[
\text{Moles of } HCl = \frac{46.9 \text{ g}}{36.46 \text{ g/mol}} \approx 1.287 \text{ moles}
\]
### Step 2: Determine the limiting reagent.
The balanced reaction is:
\[
4 HCl(aq) + MnO_2(s) \rightarrow MnCl_2(aq) + 2H_2O(l) + Cl_2(g)
\]
From the equation, the molar ratio of \(HCl\) to \(MnO_2\) is 4:1. Therefore, we need 4 moles of \(HCl\) for every mole of \(MnO_2\).
Calculate the moles of \(HCl\) needed for 0.414 moles of \(MnO_2\):
\[
\text{Moles of } HCl \text{ needed} = 0.414 \text{ moles } MnO_2 \times 4 = 1.656 \text{ moles }
\]
Since we only have approximately 1.287 moles of \(HCl\), \(HCl\) is the limiting reagent.
### Step 3: Calculate the theoretical yield of \(Cl_2\).
From the balanced equation:
\[
4 \text{ moles of } HCl \rightarrow 2 \text{ moles of } Cl_2
\]
So, the ratio of moles of \(Cl_2\) to moles of \(HCl\) is:
\[
\text{Moles of } Cl_2 = \frac{2}{4} \times (\text{moles of } HCl) = \frac{1}{2} \times 1.287 \approx 0.6435 \text{ moles of } Cl_2
\]
### Step 4: Convert moles of \(Cl_2\) to grams.
The molar mass of \(Cl_2\) is:
\[
\text{Molar mass of } Cl_2 = 2 \times 35.45 = 70.90 \text{ g/mol}
\]
The theoretical yield of \(Cl_2\):
\[
\text{Theoretical yield of } Cl_2 = 0.6435 \text{ moles} \times 70.90 \text{ g/mol} \approx 45.7 \text{ g}
\]
### Step 5: Calculate the actual yield of chlorine.
If the yield of the reaction is 84.5%, the actual yield can be calculated as:
\[
\text{Actual yield} = \text{Theoretical yield} \times \text{percentage yield}
\]
\[
\text{Actual yield} = 45.7 \text{ g} \times 0.845 \approx 38.6 \text{ g}
\]
### Final Result:
- The theoretical yield of \(Cl_2\) is approximately **45.7 g**.
- The actual yield of \(Cl_2\) is approximately **38.6 g**.