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 limiting reactant?

1. HCI

2. MnO2

Answer 1

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

2. **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.

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

4. **Compare the actual amount of HCl available:**
- We have approximately 1.287 moles of HCl available.

5. **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.