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.