Answer:
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:
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