Question

3.30 cine ma poate ajuta ofer coroana​

Answer 1

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