Question

Ma poate ajuta cineva la problema 30 ??)): va rog e urgent !

Answer 1

2,5/M = 1/22,4  ⇒   M = 56 g/mol
CxH2y + O2 ⇒xCO2 + yH2O
0,5x + 0,5y = 89,6/22,4 = 4       x + y = 8
      12x +2y =56                        6x + y = 28
                                                -------------------
                                                 5x = 20     x = 4    y = 4        C4H8   (D)

Answer 2

Notăm cu CxHy - hidrocarbura
CxHy + x+y/4) O2 = x CO2 + y/2 H2O
V c.n. = 22,4 l 
1 mol CxHy..........x*22,4 l CO2.........y/2*22,4 l H2O
0,5 moli CxHy..............a.......................b
a = 11,2x
b = 5,6y
a+b = 89,6
11,2x+5,6y = 89,6
x+0,5y = 8 => x = 8-0,5y
V = n*22,4 = m/M*22,4
M = m*22,4/V = 2,5*22,4/1 = 56 g/mol
M CxHy = 12x+y =12(8-0,5y)+y = 56
96 - 6y +y = 56
5y = 40
y = 8
x = 8-8*0,5 = 4
FM = C4H8