Questions: Question 12 (5 points) Review question #8-4: Using above balances two chemical equations from question 9 (Review question #8-1) and result from question 11 (Review question #8-3): A reaction container holds 5.33 g of P4 and 3.77 g of O2 and reaction A occurs. If enough oxygen is available then the P4O6 reacts further to undergo reaction B. If 7.12 g of P4O6 were obtained what is the percent yield (%)? 95.1 121 75.3 82.4

Solution Steps

1. Calculate moles of \( \mathrm{P}_{4} \):

   • Molar mass of \( \mathrm{P}_{4} \) = 4 × 30.97 g/mol = 123.88 g/mol
   • Moles of \( \mathrm{P}_{4} \) = \(\frac{5.33 \, \text{g}}{123.88 \, \text{g/mol}}\)

2. Write the balanced equation for the formation of \( \mathrm{P}_{4} \mathrm{O}_{6} \):

   • \( \mathrm{P}_{4} + 3 \mathrm{O}_{2} \rightarrow \mathrm{P}_{4} \mathrm{O}_{6} \)

3. Calculate moles of \( \mathrm{P}_{4} \mathrm{O}_{6} \) formed:

   • From the balanced equation, 1 mole of \( \mathrm{P}_{4} \) produces 1 mole of \( \mathrm{P}_{4} \mathrm{O}_{6} \).
   • Moles of \( \mathrm{P}_{4} \mathrm{O}_{6} \) = Moles of \( \mathrm{P}_{4} \)

4. Calculate the theoretical mass of \( \mathrm{P}_{4} \mathrm{O}_{6} \):

   • Molar mass of \( \mathrm{P}_{4} \mathrm{O}_{6} \) = 4 × 30.97 + 6 × 16.00 g/mol
   • Theoretical mass = Moles of \( \mathrm{P}_{4} \mathrm{O}_{6} \) × Molar mass of \( \mathrm{P}_{4} \mathrm{O}_{6} \)

1. Use the actual yield of \( \mathrm{P}_{4} \mathrm{O}_{6} \):

   • Actual yield = 7.12 g

2. Calculate percent yield:

   • Percent yield = \(\left(\frac{\text{Actual yield}}{\text{Theoretical yield}}\right) \times 100\%\)

1. Compare calculated percent yield with given options:
   • Check which option (95.1, 121, 75.3, 82.4) matches the calculated percent yield.

Final Answer