The reaction ch4 (g) + 2 o2 (g) ? co2 (g) + 2 h2o (g) is: the reaction ch4 (g) + 2 o2 (g) co2 (g) + 2 h2o (g) is: an exothermic
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- The mass percent of Pentane in solution is 16.49%
- The mass percent of Hexane in solution is 83.51%
<u>Explanation</u>:
- Take 1 kg basis for the vapor: 35.5 mass% pentane = 355 g pentane with 645 g hexane.
- Convert these values to mol% using their molecular weights:
Pentane: Mp = 72.15 g/mol -> 355g/72.15 g/mol = 4.92mol
Hexane: Mh = 86.18 g/mol -> 645g/86.18 g/mol = 7.48mol
Pentane mol%: yp = 4.92/(4.92+7.48) = 39.68%
Hexane mol%: yh = 100 - 39.68 = 60.32%
Pp-vap = 425 torr = 0.555atm
Ph-vap = 151 torr = 0.199atm
- From Raoult's law we know:
Pp = xp Pp - vap = yp
Pt (1)
Ph = xh Ph - vap = yh
Pt (2)
- Since it is a binary mixture we can write xh = (1 - xp) and yh = (1 - yp), therefore (2) becomes:
(1 - xp) Ph - vap = (1 - yp)
Pt (3)
- Substituting (1) into (3) we get:
(1-xp) Ph - vap = (1 - yp)
xp
Pp - vap / yp (4)
- Rearrange for xp:
xp = Ph - vap / (Pp - vap/yp - Pp - vap + Ph - vap) (5)
- Subbing in the values we find:
Pentane mol% in solution: xp = 19.08%
Hexane mol% in solution: xh = 80.92%
- Now for converting these mol% to mass%, take 1 mol basis for the solution and multiplying it by molar mass:
mp = 0.1908 mol 72.15 g/mol
= 13.766 g
mh = 0.8092 mol 86.18 g/mol
= 69.737 g
- Mass% of Pentane solution = 13.766/(13.766+69.737)
= 16.49%
- Mass% of Hexane solution = 83.51%