100% amestec .................... 34,65% N2
404 g amestec ................... mN2 = 140g N2 (rotunjit)
=> mCO2 = 404 - 140 = 264 g
PV = n.totRT , n.tot = nr. moli amestec
M.N2 = 14x2 = 28 g/mol ; M.CO2 = 12+2x16 = 44 g/mol
n1 = mN2/M.N2 = 140/28 = 5 moli N2
n2 = mCO2/M.CO2 = 264/44 = 6 moli CO2
=> n.tot = 5+6 = 11 moli amestec
=> P = n.totRT/V = 11x0,082x(273+0)/176
= 14 atm.
dupa deschiderea robinetului:
n.final = PV/RT
= 1x176/0,082(273+0) = 7,862 moli amestec ramas
daca:
11 moli amestec initiala ...... 5 moli N2 ...... 6 moli CO2
7,862 moli amestec final ..... a moli N2 ...... b moli CO2
a = 3,574 moli N2 b = 4,288 moli CO2
=> m.N2 = axM.N2 = 3,574x28 = 100,08 g
=> m.CO2 = bxM.CO2 = 4,288x44 = 188,69 g
=> m.amestec.final = 100,08 g N2 + 188,69 g CO2 = 288,76 g
