WIH)= 14,3%
D (H2) = 21
Cx Hy=?​

1. C2H6 + Br2 = C2H5Br + HBr
C2H5Br + NaOH = C2H5OH + NaBr
C2H5OH + CH3COOH ---(H2SO4)---> CH3COOC2H5 + H2O
CH3COOC2H5 + NaOH = CH3COONa + C2H5OH
CH3COONa + NaOH + t (сплав) = CH4 + Na2CO3  CH4 + t = C2H2
C2H2 + HCl = CH2=CH-Cl    nCH2=CH-Cl = [-CH2-CH(Cl)-]n

2. CH3COOCH3 ---(NaOH/H2O)---> CH3COONa + CH3OH
CH3OH + O2 ---(V2O5/t)---> HCOH + H2O
HCOH + [Ag(NH3)2]OH = Ag + HCOOH + H2O + 2NH3
HCOH ---(H2/Ni)---> CH3OH
CH3OH + C2H5COOH ---(H2SO4)---> C2H5COOCH3 + H2O
C2H5COOCH3 + NaOH (водн) = C2H5COONa + CH3OH
C2H5COONa + NaOH + t (сплав) = C2H6 + Na2CO3

3. C3H7OH ---(Cu/t)---> C2H5CHO + H2
C2H5CHO + Cu(OH)2 = C2H5COOH + Cu2O
C2H5COOH + C2H5OH ---(H2SO4)---> C2H5COOC2H5
C2H5COOC2H5 + NaOH (водн) = C2H5COONa + C2H5OH

Дано:
m(CxHyNz)=1.86г
m(CO₂)=2.64г
m(H₂O)=2.7г
D(H₂)=15.5
Найти:
формула-?
Решение:
M(CxHyNz)=15.5*2=31г/моль
M(CO₂)=44г/моль
n(C)=n(CO₂)=m/M=2.64/44=0.06моль
m(C)=M(C)*n(C)=12*0.06=0.72г
M(H₂O)=18г/моль
n(H)=2*n(H₂O)=m/M=2*2.7/18=0.3моль
m(H)=m(H)*n(H)=1*0.3=0.3г
m(CxHyNz)=m(C)+m(H)+m(N)
m(N)=m(CxHyNz)-[m(C)+m(H)]=1.86-[0.72+0.3]=0.84г
M(N)=14г/моль
n(N)=m/M=0.84/14=0.06моль
х:y:z=0.06:0.3:0.06=1:5:1
Получаем простейшую формулу: CH₅N -->CH₃NH₂ -метиламин
Проверка:
M(CH₃NH₂)=12+3+14+2=31г/моль
M(CH₃NH₂)=M(CxHyNz) -->верно
ответ: CH₃NH₂