Sure, let's go through this problem step-by-step to determine the percent yield of HCl.
### Step 1: Determine the molar masses
- For PCl[tex]\(_3\)[/tex]:
- Phosphorus (P): 30.97 g/mol
- Chlorine (Cl): 3 × 35.45 g/mol = 106.35 g/mol
- Total molar mass of PCl[tex]\(_3\)[/tex] = 30.97 g/mol + 106.35 g/mol = 137.33 g/mol
- For HCl:
- Hydrogen (H): 1.01 g/mol
- Chlorine (Cl): 35.45 g/mol
- Total molar mass of HCl = 1.01 g/mol + 35.45 g/mol = 36.46 g/mol
### Step 2: Calculate the moles of PCl[tex]\(_3\)[/tex] used in the reaction
Given:
- Mass of PCl[tex]\(_3\)[/tex] = 62.0 g
- Molar mass of PCl[tex]\(_3\)[/tex] = 137.33 g/mol
Moles of PCl[tex]\(_3\)[/tex] = [tex]\(\frac{\text{Mass}}{\text{Molar Mass}}\)[/tex] = [tex]\(\frac{62.0\text{ g}}{137.33\text{ g/mol}}\)[/tex] ≈ 0.4515 mol
### Step 3: Use the stoichiometry of the reaction to find the moles of HCl produced
The balanced chemical equation tells us that 1 mole of PCl[tex]\(_3\)[/tex] produces 3 moles of HCl.
Thus, the moles of HCl produced = 0.4515 mol of PCl[tex]\(_3\)[/tex] × 3 = 1.3545 mol
### Step 4: Calculate the theoretical yield of HCl in grams
Given:
- Moles of HCl = 1.3545 mol
- Molar mass of HCl = 36.46 g/mol
Theoretical yield = Moles × Molar Mass = 1.3545 mol × 36.46 g/mol ≈ 49.38 g
### Step 5: Calculate the percent yield
Given:
- Actual yield of HCl = 42.0 g
- Theoretical yield of HCl ≈ 49.38 g
Percent Yield = [tex]\(\frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\)[/tex] = [tex]\(\frac{42.0\text{ g}}{49.38\text{ g}} \times 100\)[/tex] ≈ 85.05%
### Conclusion
The percent yield of HCl is approximately 85.0%, so the correct answer is:
85.0%
