При сжигании 28 мл СxHy образовалось 84 мл CO2 и 0,675 г H2O. D(H2)=21. найти CxHy.

CxHy+O2=CO2+H2O; n(CO2)=84:22,4=3,75; n(H2O)=67,5:22,4=3;n(CxHy)=28:22,4=1,25                               1,25              3,75       3                                CxHy + O2 =CO2 + H2O             :1,25                                 1                3        3                                CxHy+O2=CO2+H2O                                CxHy+O2=3CO2+3H2O                                        x=3;y=6;                                                          C3H6-пропен.