Final answer:
From the stoichiometric analysis, 11.46 g of PF5 can be formed when 9.46 g of PF3 reacts with 9.42 g of XeF4, making option A the correct answer.
Step-by-step explanation:
The calculation for determining how many grams of PF5 can be formed from the given reaction involves a stoichiometric analysis. First, we need to convert the masses of PF3 and XeF4 to moles using their molar masses. The molar mass of PF3 is 31 + (3 × 19) = 88 g/mol, and the molar mass of XeF4 is 131 + (4 × 19) = 207 g/mol. With this, we convert 9.46 g of PF3 and 9.42 g of XeF4 to moles:
- 9.46 g PF3 ÷ (1 mol PF3 / 88 g PF3) = 0.1075 mol PF3
- 9.42 g XeF4 ÷ (1 mol XeF4 / 207 g XeF4) = 0.0455 mol XeF4
Since the reaction ratio is 2 moles of PF3 to 1 mole of XeF4, XeF4 is the limiting reactant. Thus, for every mole of XeF4, 2 moles of PF5 are produced, defining the produced moles of PF5 which is 2 × 0.0455 mol = 0.091 mol PF5.
The molar mass of PF5 is 31 + (5 × 19) = 126 g/mol. Using it, we convert moles of PF5 back to grams:
- 0.091 mol PF5 × (126 g PF5 / 1 mol PF5) = 11.466 g PF5