Răspuns:
CnH2n-2 + HCl---> CnH2n-1Cl
la 1mol alchina se aditioneaza 1 molHCl sau
14n-2 g alchina...........36,5gHCl
100g..............................91,25g
n=3-----> C3H4 propina cu M= 40g/mol
1mol............4mol sau
C3H4 + 4O2---> 3CO2+2H2O
40g............4x22,4l
4g............V,O2=8,96L
Vaer=100x8,96/20 l=44,8l
Explicație:mol HcL