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%