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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 limiting reactant?

1. HCI

2. MnO2

1 answer

To determine the limiting reactant in the reaction of hydrochloric acid (HCl) with manganese(IV) oxide (MnO₂), we need to follow these steps:

  1. Calculate the number of moles of reactants:

    • Molecular weight of MnO₂:
      • Manganese (Mn): 54.94 g/mol
      • Oxygen (O): 16.00 g/mol
      • MnO₂ = 54.94 + (2 × 16.00) = 54.94 + 32.00 = 86.94 g/mol

  2. Calculate moles of MnO₂: \[ \text{Moles of MnO₂} = \frac{36.1 \text{ g}}{86.94 \text{ g/mol}} \approx 0.415 \text{ mol} \]

  3. Molecular weight of HCl:

    • Hydrogen (H): 1.01 g/mol
    • Chlorine (Cl): 35.45 g/mol
    • HCl = 1.01 + 35.45 = 36.46 g/mol

  4. Calculate moles of HCl: \[ \text{Moles of HCl} = \frac{46.9 \text{ g}}{36.46 \text{ g/mol}} \approx 1.287 \text{ mol} \]

  5. Use the stoichiometry of the reaction: The balanced equation is: \[ 4 \text{ HCl} + 1 \text{ MnO₂} \rightarrow 1 \text{ MnCl₂} + 2 \text{ H₂O} + 1 \text{ Cl₂} \]

    According to the equation:

    • 4 moles of HCl react with 1 mole of MnO₂.
    • This means for every 1 mole of MnO₂, we need 4 moles of HCl.

  6. Calculate the amount of HCl required for 0.415 moles of MnO₂: \[ \text{Moles of HCl required} = 0.415 \text{ mol MnO₂} \times 4 = 1.66 \text{ mol HCl} \]

  7. Compare the actual amount of HCl available:

    • We have approximately 1.287 moles of HCl available.

  8. Determine the limiting reactant: Since we need 1.66 moles of HCl to fully react with 0.415 moles of MnO₂ but only have 1.287 moles, HCl is not enough to react with all the MnO₂.

Therefore, HCl is the limiting reactant in this reaction.