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

Answer 1

To calculate the theoretical yield of chlorine gas (Cl₂) from the reaction of hydrochloric acid (HCl) with manganese(IV) oxide (MnO₂), we first need to determine the number of moles of each reactant.

1. **Calculate moles of MnO₂:**
- The molar mass of MnO₂ is:
- Manganese (Mn): 54.94 g/mol
- Oxygen (O): 16.00 g/mol × 2 = 32.00 g/mol
- Total: 54.94 g/mol + 32.00 g/mol = 86.94 g/mol

\[
\text{Moles of MnO}_2 = \frac{\text{Mass}}{\text{Molar mass}} = \frac{36.1 \, \text{g}}{86.94 \, \text{g/mol}} \approx 0.414 \, \text{mol}
\]

2. **Calculate moles of HCl:**
- The molar mass of HCl is:
- Hydrogen (H): 1.01 g/mol
- Chlorine (Cl): 35.45 g/mol
- Total: 1.01 g/mol + 35.45 g/mol = 36.46 g/mol

\[
\text{Moles of HCl} = \frac{46.9 \, \text{g}}{36.46 \, \text{g/mol}} \approx 1.287 \, \text{mol}
\]

3. **Determine the limiting reactant:**
The balanced equation is:

\[
4 \, \text{HCl} + \text{MnO}_2 \rightarrow \text{MnCl}_2 + 2 \, \text{H}_2O + \text{Cl}_2
\]

From the balanced equation, we see that 4 moles of HCl react with 1 mole of MnO₂. This gives us the following ratios:
- For 0.414 moles of MnO₂, the required moles of HCl are:
\[
0.414 \, \text{mol MnO}_2 \times 4 \, \frac{\text{mol HCl}}{\text{mol MnO}_2} = 1.656 \, \text{mol HCl}
\]

Since we have only 1.287 moles of HCl, we can see that HCl is the limiting reactant.

4. **Calculate the theoretical yield of Cl₂:**
According to the balanced equation, 4 moles of HCl produce 1 mole of Cl₂. Thus, the moles of Cl₂ produced from 1.287 moles of HCl can be calculated as follows:

\[
\text{Moles of Cl}_2 = \frac{1.287 \, \text{mol HCl}}{4} \approx 0.32175 \, \text{mol Cl}_2
\]

5. **Convert moles of Cl₂ to grams:**
The molar mass of Cl₂ (Chlorine gas) is:
- Chlorine (Cl): 35.45 g/mol × 2 = 70.90 g/mol

Now we can find the mass of Cl₂ produced:

\[
\text{Mass of Cl}_2 = 0.32175 \, \text{mol} \times 70.90 \, \text{g/mol} \approx 22.83 \, \text{g}
\]

Thus, the theoretical yield of Cl₂ is approximately **22.83 grams**.