Question

Doi alcani omologi aflati intr-un amestec in raportul molar 1:2 formeaza prin ardere 528g CO2 si se stie ca raportul cantitatilor de O2 folosite la ardere este 0,35. Afla procentul masic de alcan inferior din amestec

Answer 1

CnH2n+2
Cn+1H2n+4
CnH2n+2 + 3n+1)/2 O2 = n CO2 + n+1 H2O
Cn+1H2n+4 + 3n+4)/2 O2 = n+1 CO2 + n+2 H2O
M CO2 = 44 g/mol
1 mol alcan 1...........44 n g CO2
x moli alcan 1...............a g CO2
44nx = a
1 mol alcan 2...........44n+44 g CO2
y moli alcan 2.....................b
b = 44ny+44y
a + b = 528
44nx+44ny+44y = 528
x/y = 1/2 
y = 2x
44nx+44n*2x+44*2x = 528
132nx+88x=528
1,5nx+x = 6
1 mol alcan 1........3n+1/2 moli O2
x moli alcan 1................c 
c = 1,5nx+0,5x
1 mol alcan 2...........(3n+4)/2 moli O2
y moli alcan 2...................d
d = 1,5ny+2y = 3nx+4x
(1,5nx+0,5x)/(3nx+4x) = 0,35
1,5nx+0,5x = 1,05nx + 1,4x
0,45nx = 0,9x
0,45n = 0,9
n = 2
=> C2H6 și C3H8
1,5nx+x = 6
1,5*2*x+x = 6
3x+x = 6
x = 1,5 moli C2H6
y = 2x = 1,5 * 2 = 3 moli C3H8
M C2H6 = 30 g/mol
M C3H8 = 44 g/mol
m amestec = n C2H6 * M + n C3H8 * M = 1,5*30 + 2*44 = 133 g
% C2H6 = m C2H6/m amestec * 100 = 45/133 * 100 = 33,83%