Consider the reaction MnO2 + 4HCl ? MnCl2 + Cl2 + H2O If 2.86 mol of MnO2 and 96.2 g of HCl react, which reagent will be used up first?
The Correct Answer and Explanation is :
Answer:
Hydrochloric acid (HCl) is the limiting reagent.
Explanation:
To determine which reagent is the limiting reagent, follow these steps:
Step 1: Write the balanced chemical equation
[
\text{MnO}_2 + 4\text{HCl} \rightarrow \text{MnCl}_2 + \text{Cl}_2 + 2\text{H}_2O
]
This equation shows that 1 mole of MnO₂ reacts with 4 moles of HCl.
Step 2: Calculate the moles of each reactant
- Moles of MnO₂:
Given that we have 2.86 moles of MnO₂, this value remains unchanged. - Moles of HCl:
Given mass = 96.2 g, and the molar mass of HCl = 1.01 (H) + 35.45 (Cl) = 36.46 g/mol, [
\text{Moles of HCl} = \frac{96.2 \text{ g}}{36.46 \text{ g/mol}} = 2.64 \text{ moles}
]
Step 3: Determine the limiting reagent
From the balanced equation:
- 1 mole of MnO₂ requires 4 moles of HCl.
- 2.86 moles of MnO₂ would require: [
2.86 \times 4 = 11.44 \text{ moles of HCl}
]
However, we only have 2.64 moles of HCl, which is much less than the required 11.44 moles. This means that HCl will be used up first, making it the limiting reagent.
Step 4: Conclusion
Since HCl is consumed first, the reaction will stop once all of it is reacted, even though there is still some MnO₂ left. This means that HCl is the limiting reagent, and MnO₂ is the excess reagent.