Question

Ce volum de gaze la presiune normala si 527°C ( H2O e in stare de vapori ) se formeaza prin arderea a 11.2 dm ^3 etan (c.n) ??

Answer 1

11,2 dm3 .................. a kmoli ...b kmoli

C2H6 + 7/2O2 --> 2CO2 + 3H2O

22,4 ............................2  ........... 3

=> a = 2x11,2/22,4 = 1 kmol CO2

=> b = 3x11,2/22,4 = 1,5 kmoli H2O gazoasa

stim ca PV = nRT

n = a + b = 1,5+1 = 2,5 kmoli

=> 1V = 2,5x0,82x(273+527)

=> V = 164 dm3 gaze