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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?

1 answer

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.