Just when I thought I\'d gotten the hang of these gas stoichiometryproblems, I r

Question

Just when I thought I'd gotten the hang of these gas stoichiometryproblems, I ran into this.
The problem and my attempt at solving it are shown below.Unfortunately, my answers did not match the ones at the back of thebook, so I need your help.

Problem:
When gaseous F2 and solid I2 are heated tohigh temeratures, the I2 sublimes and gaseous iodineheptafluoride forms. If 350 torr of F2 and 2.50g ofsolid I2 are put into a 2.50 L container at 250K and thecontainer is heated to 550K, what is the final pressure? What isthe partial pressure of I2 gas?

My solution attempt:

I2(s) + 7F2(g) --> 2IF7(g)

Initially,
nF2 = PV/RT = (350 x 1/760)(2.50)/(0.0821)(250) = 0.0561mol F2(g)
nI2 = 2.50/126.9 = 0.0197 mol I2(s)

The F2 gas is the limiting reactant. Therefore,

0.0561 mol F2 x (1 mol I2 / 7 molF2) = 8.01x10-3 mol I2 consumed inreaction.
0.0561 mol F2 x (2 mol IF7 / 7 molF2) = 0.0160 mol IF7 produced.

nI2 remaining = 0.0197 - 8.01x10-3 = 0.0117mol I2(g) remaining.

Pfinal = ntotalRT/V = (nI2remaining + nIF7)RT/V = (0.0117 +0.0160)(0.0821)(550)/2.50 = 0.500 atm.
PI2 = nI2RT/V = (0.0117)(0.0821)(550)/2.50 =0.211 atm.

Apparently, my answers are wrong.
The correct answers are: Pfinal = 0.323 atm andPI2 = 0.0332 atm.
Could someone please help me out?

Thank you.

Explanation / Answer

Sorry, I just realized that I made a stupid mistake and I didn'tknow how to delete my post. The molar mass of I2 is 2 x 126.9 g/mol. Once thatcorrection is made, everything works out correctly. Thanks anyway:)

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