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Qube
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Homework Statement
When fluorine and solid iodine are heated at 550. K, the iodine completely sublimes and gaseous iodine heptafluoride forms. If 350. torr of fluorine gas and 2.50 g of solid iodine are put into a 2.50 L container at 250. K and the container is heated to 550. K, what is the final pressure in the flask?
Homework Equations
5F2 (g) + I2 (s) --> 2IF5(g)
The Attempt at a Solution
We have 350/760 atm of fluorine gas. Using the ideal gas law, P = nRT/V, we find taht we have 0.056 moles of F.
We also have 0.009 moles of solid iodine (2.5 grams / twice the molecular mass of elemental iodine).
This yields 0.0197 moles of product. Iodine is limiting and fluorine is in excess. The final pressure, by Dalton's Law of Partial Pressures, is the pressure of the product and the pressure of the remaining reactants.
We have 0.0197 * (5/2) moles of fluorine used. This means we have 0.056 minus 0.0475 moles of fluorine remaining. This means there are 0.085 moles of fluorine remaining. The pressure of this remaining fluorine is:
P = nRT/V = 0.085(0.0821)(550)/2.5 = 0.15 atm.
The pressure of the product is also nRT/V = 0.0197 * 0.0821 * 550 / 2.5 = 0.355 atm.
Summing these two pressures, we get 0.50 atm of pressure. However, this isn't an answer choice. The closest choice is 0.534 atm, leading me to believe there is an error in my calculations, and not a rounding one at that. Is there any flaw in my line of reasoning?