1) C6H14 -> C6H6 + 4H2
C6H6 + CH3-CHCl-CH3 -> C6H5-CH(CH3)2
C6H5-CH(CH3)2 +HOH -> C6H5OH +CH3-CH2-CH3
C6H5OH+3Br2 -> C6H2Br3OH +3HBr
12 8,4
5) CnH2n+2O -> CnH2n + H2O
14n+18 14n
Решим пропорцию и найдем "n"
n=3
Ответ: C3H8O
4)2C2H5OH +2Na -> 2C2H5ONa +H2
n(C2H5OH)=13,8/46=0,3;
n(Na)=9,2/23=0,4; Na-в избытке!
n(C2H5ONa)=0,3; m(C2H5ONa)=0,3*68=20,4;