Final answer:
To find the partial pressure of IF5, we calculate the moles of reactants, identify the limiting reagent, calculate moles of IF5 formed, and use the ideal gas law to determine the partial pressure of IF5, which is 0.694 atm.
Step-by-step explanation:
Calculating Partial Pressure
To find the partial pressure of IF5 in the flask after the reaction between I2 and F2, we need to perform stoichiometric calculations to determine which reactant is the limiting reagent and the amount of IF5 formed. However, the reaction between I2 and F2 to form IF5 is not provided in the initial question. Assuming the reaction is I2(g) + 5F2(g) → 2IF5(g), we can proceed. First, we calculate the moles of I2 and F2 present in the flask:
- Moles of I2: (10.0 g)/(253.8 g/mol) = 0.0394 mol
- Moles of F2: (10.0 g)/(38.0 g/mol) = 0.2632 mol
Since the mole ratio of I2:F2 is 1:5, F2 is the limiting reagent. The reaction consumes all F2 and produces IF5:
- Moles of IF5 produced: (0.2632 mol F2)*(2 mol IF5 / 5 mol F2) = 0.1053 mol IF5
Now we use the ideal gas law (PV=nRT) to find the partial pressure of IF5:
- P = (nRT)/V
- P = [(0.1053 mol)*(0.0821 L·atm/(mol·K))*(398.15 K)]/(5.00 L)
- P = 0.694 atm (rounded to three significant figures)
The partial pressure of IF5 is therefore 0.694 atm at 125 °C in the 5.00-liter flask.