Hi, I\' been trying to solve this problem...for mixture of gases A 2.42 gram sam

Question

Hi, I' been trying to solve this problem...for mixture of gases
A 2.42 gram sample of PCl 5 was placed into an evacuated 2.00L flask and allowed to partially decompose at 250C . According tothis equation: PCl 5 (g) -> PCl 3(g) + Cl2(g) The total pressure in the flask after partial decompositionwas 359 torr .calculate the mole fraction of each gas in theflask
Thanks!!
A 2.42 gram sample of PCl 5 was placed into an evacuated 2.00L flask and allowed to partially decompose at 250C . According tothis equation: PCl 5 (g) -> PCl 3(g) + Cl2(g) The total pressure in the flask after partial decompositionwas 359 torr .calculate the mole fraction of each gas in theflask
Thanks!!

Explanation / Answer

WE Know that :      The given equation is :       PCl 5 (g) -> PCl 3(g) +Cl2(g)       According to ideal gas equation:      PV = nRT      359 torr / 760 torr x 2.0L = n x 0.0821 atm-L / mol-K x 298 K           numberof moles = 0.03861 moles         number of molesof PCl5 = 2.42 g / 208.5 g/mol                                                  = 0.0116 moles          According tothe equation :           PCl 5(g) -> PCl 3(g) + Cl2(g)   I       0.0116        0             0     C       -x             +x          +x     E     0.0116 -x        +x      + x      Total moles = 0.0116-x +x+x = 0.03861                           x = 0.02700 moles         mole fractionof PCl5 = 0.01541 moles / 0.03861moles                                           = 0.399         mole fractionof PCl3 = 0.02700 moles / 0.03861moles                                            = 0.699   = mole fraction of Cl2

Related Questions

Navigate

• Browse (All)
• Browse H
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.