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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? If the yield of the reaction is 84.5% what is the actual yield of chlorine?

1 answer

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.