$n_X= 0,3 mol$ 

Gọi a, b, c là mol propan, etilen, axetilen.

=> $a+b+c=0,3$                           (1)

Bảo toàn $\pi$: $b+2c=0,326$   (2) 

$C_2H_2+ 2AgNO_3+ 2NH_3 \to Ag_2C_2 \downarrow+ 2NH_4NO_3$

=> $240c=24,24$                         (3)

(1)(2)(3) => $a=0,075; b=0,124; c=0,101$

$\%V_{C_3H_8}= \frac{0,075.100}{0,3}= 25\%$

$\%V_{C_2H_4}= \frac{0,124.100}{0,3}= 41,33\%$

$\%V_{C_2H_2}= 33,67\%$

nX=6,72/ 22,4=0,3 mol

Gọi x, y, z lần lượt là nC3H8, nC3H6, nC3H4

PTHH: C2H4+Br2-->C2H4Br2

             y        y

C2H2+2Br2-->C2H2Br4

 z          2z

C2H2+2AgNO3+2NH3-->C2Ag2+2NH4NO3

0,101                                  0,101

m kết tủa=mC2Ag2=24,24 g

-->nC2Ag2=$\frac{24,24}{12. 2+108. 2}$ =0,101 mol=nC2H2(PTHH)

Ta có: nBr2=y+2z=0,326

                 =y+2. 101=0,326

-->y=0,124 mol=nC2H4

-->nPropan=0,3-0,101-0,124=0,075 mol

%V C3H8=$\frac{0,075 .22,4. 100}{6,72}$ =25%

%V C2H4=$\frac{0,124. 22,4. 100}{6,72}$ =41,33%

-->%V C2H2=33,67%