Phần 1:

CnH2n+1OH+Na→CnH2n+1ONa+12H2CmH2m+1COOH+Na→CmH2m+1COONa+12H2

Theo phương trình có nX(P1)=2nH2=2.0,07=0,14mol

Phần 2:

CnH2n+2O+3n2O2→t°nCO2+(n+1)H2O

Cm+1H2(m+1)O2+3n−22O2  →t°    (m+1)CO2+(m+1)H2O

Theo phương trình: nCnH2n+2O(P1=P2)=nH2O−nCO2=0,28−0,22=0,06mol

⇒nCmH2m+1COOH=0,14−0,06=0,08mol

CnH2n+2O   +  3n2O2→t°   nCO2   +   (n+1)H2O0,06                                        0,06n               0,06.(n+1)  mol

Cm+1H2(m+1)O2+3n−22O2→t°(m+1)CO2+(m+1)H2O0,08                                                     0,08(m+1)         0,08(m+1)   mol

∑nCO2=0,06n+(0,08(m+1)=0,22⇔0,06n+0,08m=0,14⇔6n+8m=14(1)⇒8m<14⇒m<1,75

Vì m nguyên, mm ≥0⇒m=0;1

Xét bảng:

Vậy rượu CnH2n+1OH(n≥1)là CH₃OH

axit CmH2m+1COOH(m≥0)là CH₃COOH