a)

nCO2=nC=11,648/22,4=0,52mol

→nO=0,52.2=1,04mol

nH=2nH2O=14,04/18.2=1,56mol

→nO=1,56/2=0,78mol

nO2 dùng=14,56/22,4=0,65mol

→nO dùng=0,65.2=1,3mol

BTO

nO trong A=1,04+0,78-1,3=0,52mol

C:H:O=0,52:1,56:0,52=2:3:2

Công thức đơn giản: (CH3O)n

MA=31.2=62(đvc)

→(CH3O)n=62

→31n=62

→n=2

CTHH A: C2H6O2

b) nCO2=nC=5,824/22,4=0,26mol

→nO=0,26.2=0,52mol

nH2O=nO=7,02/18=0,39mol

→nH=0,39.2=0,78mol

nO dùng=7,28/22,4.2=0,65mol

BTO: nO trong B=0,52+0,39-0,65=0,26mol

C:H:O=0,26:0,78:0,26=1:3:1

Công thức đơn giản: (CH3O)n

MB=31.2=62(đvc)

→(CH3O)n=62

→31n=62

→n=2

CTHH B=C2H6O2

a)

Sơ đồ phản ứng:

\(A + {O_2}\xrightarrow{{{t^o}}}C{O_2} + {H_2}O\)

Ta có:

\({n_{C{O_2}}} = \frac{{11,648}}{{22,4}} = 0,52{\text{ mol = }}{{\text{n}}_C}\)

\({n_{{H_2}O}} = \frac{{14,04}}{{18}} = 0,78{\text{ mol}} \to {{\text{n}}_H} = 2{n_{{H_2}O}} = 1,56{\text{ mol}}\)

\({n_{{O_2}}} = \frac{{14,56}}{{22,4}} = 0,65{\text{ mol}}\)

Bảo toàn \(O\)

\({n_{O{\text{ trong A}}}} = 2{n_{C{O_2}}} + {n_{{H_2}O}} - 2{n_{{O_2}}} = 0,52{\text{ mol}}\)

\( \to n_C:n_H:n_O=0,52:1,56:0,52=1:3:1\)

Vậy \(A\) có dạng \((CH_3O)_n\)

\( \to {M_A} = (12 + 3 + 16)n = 31n = 31{M_{{H_2}}} = 31.2 = 62\)

\( \to n=2 \to A: C_2H_6O_2\)

b)

Sơ đồ phản ứng:

\(B + {O_2}\xrightarrow{{{t^o}}}C{O_2} + {H_2}O\)

Ta có:

\({n_{C{O_2}}} = \frac{{5,824}}{{22,4}} = 0,26{\text{ mol = }}{{\text{n}}_C}\)

\({n_{{H_2}O}} = \frac{{7,02}}{{18}} = 0,39{\text{ mol}} \to {{\text{n}}_H} = 2{n_{{H_2}O}} = 0,78{\text{ mol}}\)

\({n_{{O_2}}} = \frac{{7,28}}{{22,4}} = 0,325{\text{ mol}}\)

Bảo toàn \(O\)

\({n_{O{\text{ trong B}}}} = 2{n_{C{O_2}}} + {n_{{H_2}O}} - 2{n_{{O_2}}} = 0,26{\text{ mol}}\)

\( \to n_C:n_H:n_O=0,26:0,78:0,26=1:3:1\)

Vậy \(B\) có dạng \((CH_3O)_n\)

\( \to {M_B} = (12 + 3 + 16)n = 31n = 31{M_{{H_2}}} = 31.2 = 62\)

\( \to n=2 \to B: C_2H_6O_2\)