When gaseous f2 and solid i2 are heated to high temperatures, the i2 sublimes and gaseous iodine heptafluoride forms.3.30 × 102 torr of f2 and 4.40 g of solid i2 are put into a 2.50 l container at 2.50 × 102 k and the container is heated to 5.50 × 10^2 k.

(a) what is the final pressure?
ptotal =

(b) what is the partial pressure of i2 gas?
pi2 = i2

a)

b)

Explanation:

Hello,

At first, we state the undergoing chemical reaction as:

Now, from the beginning we know that there are 4.40 g of solid iodine and the reacting fluorine is at 330 torr, thus, one finds the reacting moles  as follows:

and identify the limiting reagent based on the theoretical moles of solid iodine that completely react with 0.0530 moles of gaseous fluorine as:

Now, as long as there are 0.0173 moles of solid iodine and the reacting amount is smaller, one states that the fluorine is the limiting reagent and there are 0.0173-0.00756= 0.00974 moles of unreacted solid iodine which are subsequently sublimed.

Under these conditions, the stoichiometry allows us to compute the resulting moles of iodine heptafluoride as:

a) Now, to compute the final pressure, one must include both the gaseous moles of iodine heptafluoride and the sublimed iodine by using the ideal gas equation but now at 550K, as follows:

b) Finally, for the partial pressure of the sublimed iodine gas, one just takes the unreacted moles which were sublimed due to the heat to compute the pressure as:

Best regards.