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?
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?
Answers
GPT-4o mini
To calculate the theoretical yield of chlorine gas (\(Cl_2\)) from the reaction between hydrochloric acid (\(HCl\)) and manganese(IV) oxide (\(MnO_2\)), we first need to determine the limiting reactant by calculating the number of moles of each reactant.
**Step 1: Calculate the moles of \(MnO_2\) and \(HCl\).**
1. **Molar mass of \(MnO_2\):**
- Manganese (Mn): 54.94 g/mol
- Oxygen (O): 16.00 g/mol
- Molar mass of \(MnO_2 = 54.94 + 2 \times 16.00 = 54.94 + 32.00 = 86.94 \, \text{g/mol}\)
\[
\text{Moles of } MnO_2 = \frac{\text{Mass}}{\text{Molar mass}} = \frac{36.1 \, \text{g}}{86.94 \, \text{g/mol}} \approx 0.415 \, \text{mol}
\]
2. **Molar mass of \(HCl\):**
- Hydrogen (H): 1.01 g/mol
- Chlorine (Cl): 35.45 g/mol
- Molar mass of \(HCl = 1.01 + 35.45 = 36.46 \, \text{g/mol}\)
\[
\text{Moles of } HCl = \frac{46.9 \, \text{g}}{36.46 \, \text{g/mol}} \approx 1.287 \, \text{mol}
\]
**Step 2: Determine the limiting reactant using the stoichiometry of the reaction.**
The balanced equation is:
\[
4 \, HCl + MnO_2 \rightarrow MnCl_2 + 2 \, H_2O + Cl_2
\]
From the equation, we see that:
- 4 moles of \(HCl\) react with 1 mole of \(MnO_2\).
Now we need to find how many moles of \(HCl\) are required for the available moles of \(MnO_2\):
- For 0.415 moles of \(MnO_2\), the required moles of \(HCl\) are:
\[
0.415 \, \text{mol} \, MnO_2 \times 4 \, \frac{\text{mol} \, HCl}{1 \, \text{mol} \, MnO_2} = 1.66 \, \text{mol} \, HCl
\]
Since we only have 1.287 moles of \(HCl\), \(HCl\) is the limiting reactant.
**Step 3: Calculate the theoretical yield of \(Cl_2\).**
From the balanced equation, we see that:
- 4 moles of \(HCl\) produce 1 mole of \(Cl_2\).
The moles of \(Cl_2\) produced can be calculated as:
\[
\text{Moles of } Cl_2 = \frac{1.287 \, \text{mol} \, HCl}{4} = 0.32175 \, \text{mol} \, Cl_2
\]
**Step 4: Calculate the mass of \(Cl_2\) produced.**
The molar mass of \(Cl_2\):
- Chlorine (Cl): 35.45 g/mol
- Molar mass of \(Cl_2 = 2 \times 35.45 = 70.90 \, \text{g/mol}\)
Finally, we can find the mass of \(Cl_2\):
\[
\text{Mass of } Cl_2 = \text{Moles} \times \text{Molar mass} = 0.32175 \, \text{mol} \times 70.90 \, \text{g/mol} \approx 22.82 \, \text{g}
\]
**Theoretical yield of \(Cl_2\) is approximately 22.82 g.**
**Step 1: Calculate the moles of \(MnO_2\) and \(HCl\).**
1. **Molar mass of \(MnO_2\):**
- Manganese (Mn): 54.94 g/mol
- Oxygen (O): 16.00 g/mol
- Molar mass of \(MnO_2 = 54.94 + 2 \times 16.00 = 54.94 + 32.00 = 86.94 \, \text{g/mol}\)
\[
\text{Moles of } MnO_2 = \frac{\text{Mass}}{\text{Molar mass}} = \frac{36.1 \, \text{g}}{86.94 \, \text{g/mol}} \approx 0.415 \, \text{mol}
\]
2. **Molar mass of \(HCl\):**
- Hydrogen (H): 1.01 g/mol
- Chlorine (Cl): 35.45 g/mol
- Molar mass of \(HCl = 1.01 + 35.45 = 36.46 \, \text{g/mol}\)
\[
\text{Moles of } HCl = \frac{46.9 \, \text{g}}{36.46 \, \text{g/mol}} \approx 1.287 \, \text{mol}
\]
**Step 2: Determine the limiting reactant using the stoichiometry of the reaction.**
The balanced equation is:
\[
4 \, HCl + MnO_2 \rightarrow MnCl_2 + 2 \, H_2O + Cl_2
\]
From the equation, we see that:
- 4 moles of \(HCl\) react with 1 mole of \(MnO_2\).
Now we need to find how many moles of \(HCl\) are required for the available moles of \(MnO_2\):
- For 0.415 moles of \(MnO_2\), the required moles of \(HCl\) are:
\[
0.415 \, \text{mol} \, MnO_2 \times 4 \, \frac{\text{mol} \, HCl}{1 \, \text{mol} \, MnO_2} = 1.66 \, \text{mol} \, HCl
\]
Since we only have 1.287 moles of \(HCl\), \(HCl\) is the limiting reactant.
**Step 3: Calculate the theoretical yield of \(Cl_2\).**
From the balanced equation, we see that:
- 4 moles of \(HCl\) produce 1 mole of \(Cl_2\).
The moles of \(Cl_2\) produced can be calculated as:
\[
\text{Moles of } Cl_2 = \frac{1.287 \, \text{mol} \, HCl}{4} = 0.32175 \, \text{mol} \, Cl_2
\]
**Step 4: Calculate the mass of \(Cl_2\) produced.**
The molar mass of \(Cl_2\):
- Chlorine (Cl): 35.45 g/mol
- Molar mass of \(Cl_2 = 2 \times 35.45 = 70.90 \, \text{g/mol}\)
Finally, we can find the mass of \(Cl_2\):
\[
\text{Mass of } Cl_2 = \text{Moles} \times \text{Molar mass} = 0.32175 \, \text{mol} \times 70.90 \, \text{g/mol} \approx 22.82 \, \text{g}
\]
**Theoretical yield of \(Cl_2\) is approximately 22.82 g.**