Answer:
a) [tex]P_T=0.447atm=340.02 torr[/tex]
b) [tex]P_{I_2}=0.176atm=133.54 torr[/tex]
Explanation:
Hello,
At first, we state the undergoing chemical reaction as:
[tex]7F_2(g)+I_2(s)-->2IF_7(g)[/tex]
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:
[tex]n_{I_2}=4.40gI_2*\frac{1molI_2}{254gI_2}=0.0173molI_2[/tex]
[tex]n_{F_2}=\frac{330 torr*\frac{1atm}{760 torr}*2.50L}{0.082\frac{atm*L}{mol*K} *K} =0.053molF_2[/tex]
and identify the limiting reagent based on the theoretical moles of solid iodine that completely react with 0.0530 moles of gaseous fluorine as:
[tex]0.0530molF_2*\frac{1molI_2}{7molF_2}=0.00756molI_2[/tex]
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:
[tex]n_{IF_7}=0.053molF_2*\frac{2molIF_7}{7molF_2}=0.0151molIF_7[/tex]
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:
[tex]n_T=0.0151mol+0.00974mol=0.0248mol\\ P_T=\frac{nRT}{V}=\frac{0.0248mol*0.082\frac{atm*L}{mol*K}*550K}{2.50L} =0.447atm=340.02 torr[/tex]
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:
[tex]P_{I_2}=\frac{n_{I_2}RT}{V}=\frac{0.00974mol*0.082\frac{atm*L}{mol*K}*550K}{2.50L} =0.176atm=133.54 torr[/tex]
Best regards.
