Gaseous iodine pentafluoride, IF5, can be prepared by the reaction of solid iodi

Question

Gaseous iodine pentafluoride, IF5, can be prepared by the reaction of solid iodine and gaseous fluorine:

I2(s)+5F2(g)?2IF5(g)

A 4.50?L flask containing 9.00g I2 is charged with 9.00g F2, and the reaction proceeds until one of the reagents is completely consumed. After the reaction is complete, the temperature in the flask is 125 ?C.

Part A

What is the partial pressure of IF5 in the flask?

Part B

What is the mole fraction of IF5 in the flask?

??

Part C

What is the total mass of reactants and products in the flask?

Explanation / Answer

Okay, well first you should convert the grams I2 and F2 into moles.

F2 = 38.0g/mol and I2 = ~254g/mol

So you have 0.263 moles F2 and 0.0394 moles I2

I2 is the limiting reacting, do you see? For every I2, you need 5 times as much, F2, and you have more than five times F2 than I2.

0.0394 x 5 (mole ratio for F2) = .197 moles F2 used up

leaves 0.066moles of F2. And creates 0.0394 x 2 = .0788 moles IF5

Mole fraction would be .0788/(0.066+.0788) (moles IF5/total moles)

You would then calculate the total pressure of all the gas (just combine the moles of both gas PV = nRT does not take molecular mass into account) And then multiply that by the mole fraction.

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