Answer:
The answer is "82.2 torr"
Explanation:
moles of Xe:
[tex]= \frac{0.06}{131.293} \\\\ =0.00045699313 \ \mol[/tex]
moles of [tex]F_2[/tex]:
[tex]= \frac{0.0274}{38} \\\\= 0.00072105263\ \ mol[/tex]
moles of produced [tex]XeF_2:[/tex]
[tex]= 0.00024[/tex]
moles of left [tex]Xe[/tex]:
[tex]= 0.00021[/tex]
Calculate the Pressure:
[tex]= \frac{(0.08206\times 0.00024 \times 293)}{(.1)} + \frac{(0.08206\times 0.00021 \times 293)}{(.1)} \\\\= 0.10819611 \ \ atm \\\\ = 0.10819611 \times 760 \\\\ = 82.2 \ \ torr[/tex]