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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

1 answer

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.