Consider the reaction MnO2(s) 4HCl(aq) ? MnCl2(aq) Cl2(g) 2H2O(l). If 74.8 g MnO2 and 48.2 g HCl react, what is the theoretical yield of Cl2 closest to?
The Correct Answer and Explanation is:
To find the theoretical yield of Cl₂, we follow these steps:
Balanced Chemical Equation:
MnO2(s)+4HCl(aq)→MnCl2(aq)+Cl2(g)+2H2O(l)\text{MnO}_2 (s) + 4\text{HCl} (aq) \rightarrow \text{MnCl}_2 (aq) + \text{Cl}_2 (g) + 2\text{H}_2O (l)
Step 1: Calculate molar masses
- MnO₂:
Mn = 54.94 g/mol, O = 16.00 g/mol
→ MnO₂ = 54.94 + 2×16.00 = 86.94 g/mol - HCl:
H = 1.01, Cl = 35.45
→ HCl = 1.01 + 35.45 = 36.46 g/mol
Step 2: Convert mass to moles
- Moles of MnO₂: 74.8 g86.94 g/mol=0.860moles\frac{74.8 \text{ g}}{86.94 \text{ g/mol}} = 0.860 moles
- Moles of HCl: 48.2 g36.46 g/mol=1.32moles\frac{48.2 \text{ g}}{36.46 \text{ g/mol}} = 1.32 moles
Step 3: Identify limiting reactant
From the balanced equation, 1 mol MnO₂ reacts with 4 mol HCl.
So, 0.860 mol MnO₂ would require: 0.860×4=3.44 mol HCl0.860 \times 4 = 3.44 \text{ mol HCl}
But we only have 1.32 mol HCl, so HCl is the limiting reactant.
Step 4: Determine moles of Cl₂ produced
From the balanced equation, 4 mol HCl produces 1 mol Cl₂.
So, 1.32 mol HCl will produce: 1.324=0.330 mol Cl2\frac{1.32}{4} = 0.330 \text{ mol Cl}_2
Step 5: Convert moles of Cl₂ to grams
Molar mass of Cl₂ = 2 × 35.45 = 70.90 g/mol 0.330×70.90=23.4 g Cl20.330 \times 70.90 = \boxed{23.4 \text{ g Cl}_2}
✅ Final Answer: 23.4 g Cl₂
Explanation
To calculate the theoretical yield of chlorine gas (Cl₂), we begin by analyzing the balanced chemical equation: MnO2+4HCl→MnCl2+Cl2+2H2O\text{MnO}_2 + 4\text{HCl} \rightarrow \text{MnCl}_2 + \text{Cl}_2 + 2\text{H}_2O
This tells us the molar ratio: 1 mole of MnO₂ reacts with 4 moles of HCl to produce 1 mole of Cl₂. The goal is to determine the maximum mass (theoretical yield) of Cl₂ that can be formed from 74.8 g of MnO₂ and 48.2 g of HCl.
We convert grams of each reactant into moles using their molar masses. MnO₂ has a molar mass of 86.94 g/mol, so 74.8 g corresponds to 0.860 mol. HCl has a molar mass of 36.46 g/mol, so 48.2 g gives 1.32 mol. Next, we determine the limiting reactant by comparing mole ratios. Since 0.860 mol MnO₂ would need 3.44 mol HCl, but only 1.32 mol HCl is available, HCl is the limiting reactant—it will run out first and limit how much Cl₂ can be formed.
Using stoichiometry, 4 mol HCl produces 1 mol Cl₂. So 1.32 mol HCl produces 0.330 mol Cl₂. Finally, converting this to grams using Cl₂’s molar mass (70.90 g/mol), we find the theoretical yield is: 0.330×70.90=23.4 grams0.330 \times 70.90 = \boxed{23.4 \text{ grams}}
This is the maximum amount of chlorine gas that can be formed from the given amounts of MnO₂ and HCl, assuming complete reaction and no losses.
