Answer
Pfinal = 0.323 atm and PI[sub]2[/sub] = 0.0332 atm
Work Step by Step
I2(s) + 7F2(g) --> 2IF7(g)
Initially,
nF[sub]2[/sub] = PV/RT = (350 x 1/760)(2.50)/(0.0821)(250) = 0.0561 mol F2(g)
nI[sub]2[/sub] = 2.50/126.9*2 = 0.0098 mol I2(s)
The F2 gas is the limiting reactant. Therefore,
0.0561 mol F2 x (1 mol I2 / 7 mol F2) = 8.01x10-3 mol I2 consumed in reaction.
0.0561 mol F2 x (2 mol IF7 / 7 mol F2) = 0.0160 mol IF7 produced.
nI[sub]2[/sub] remaining = 0.0098 - 8.01x10-3 = 0.0018 mol I2(g) remaining.
Pfinal = ntotalRT/V = (nI[sub]2 remaining[/sub] + nIF[sub]7[/sub])RT/V = (0.0117 + 0.0160)(0.0821)(550)/2.50 = 0.323 atm.
PI[sub]2[/sub] = nI[sub]2[/sub]RT/V = (0.0018)(0.0821)(550)/2.50 = 0.0332 atm.