(CH3)2CO + H2 = CH3-CH(OH)-CH3
n(ац) = 116/58 = 2 моль
n(H2) = 22,4*2 = 44,8 л
если до спирта
MgO+2HCl=MgCl2+H2O
п(MgO)=1,5/40=0,0375
по уравнению реакции: n(HCl)= 0,075
m(вещества)=0,075*36,5=2,74
m(раствора)= 18,25г, тогда V=18,25/1,073=17
2CH4=C2H2+3H2
3C2H2=C6H6
C6H6+Cl2=C6H5Cl+HCl
C6H5Cl+NaOH=C6H5OH+NaCl
C6H5OH+3Br2=C6H2Br3OH+3HBr
б)
CH4+Cl2=CH3Cl+HCl
2CH3Cl+2Na=C2H6+2NaCl
C2H6+Cl2=C2H5Cl+HCl
C2H5Cl+CH3Cl+2Na=C3H8+2NaCl
C3H8+Cl2=C3H7Cl+HCl
2C3H7Cl+2Na=C6H14+2NaCl
C6H14=C6H6+4H2
C6H6+Cl2=C6H5Cl+HCl
C6H5Cl+NaOH=C6H5OH+NaCl
C6H5OH+3HNO3=C6H2(NO2)3OH+3H2O