a.Bảo toàn C: \(n_C=n_{CO_2}=\dfrac{4,48}{22,4}=0,2mol\)

Bảo toàn H: \(n_H=2.n_{H_2O}=2.\dfrac{3,6}{18}=0,4mol\)

\(n_O=\dfrac{6-\left(0,2.12+0,4.1\right)}{16}=0,2mol\)

=> A gồm C,H và O

\(CTPT:C_xH_yO_z\)

\(x:y:z=0,2:0,4:0,2=2:4:2\)

\(CTĐG:\left(C_2H_4O_2\right)n=60\)

             \(\Leftrightarrow n=1\)

Vậy CTPT A: \(C_2H_4O_2\) hay \(CH_3COOH\)

b.\(n_{CO_2}=\dfrac{4,48}{22,4}=0,2mol\)

\(Ca\left(OH\right)_2+CO_2\rightarrow CaCO_3+H_2O\)

                       0,2          0,2                ( mol )

\(m_{CaCO_3}=0,2.100=20g\)

\(Đặt:n_{CO_2}=a\left(mol\right),n_{H_2O}=b\left(mol\right)\)

\(BTKL:\\ m_X+m_{O_2}=m_{CO_2}+m_{H_2O}\\ \Rightarrow1.88+\dfrac{1.904}{22.4}\cdot32=44a+18b\)

\(\Rightarrow44a+18b=4.6\left(1\right)\)

\(\dfrac{m_{CO_2}}{m_{H_2O}}=\dfrac{88}{27}\Leftrightarrow\dfrac{44a}{18b}=\dfrac{88}{27}\Leftrightarrow\dfrac{a}{b}=\dfrac{4}{3}\left(2\right)\)

\(\left(1\right),\left(2\right)\Leftrightarrow\left\{{}\begin{matrix}a=0.08\\b=0.06\end{matrix}\right.\)

\(m_O=1.88-0.08\cdot12-0.06\cdot2=0.8\left(g\right)\\ n_O=\dfrac{0.8}{16}=0.05\left(mol\right)\)

\(Đặt:CTPT:C_xH_yO_z\)

\(x:y:z=0.08:0.12:0.05=8:12:5\)

\(CTPT:C_8H_{12}O_5\)

\(\dfrac{m_{CO_2}}{m_{H_2O}} = \dfrac{11}{6}\Rightarrow \dfrac{n_{CO_2}}{n_{H_2O}} = \dfrac{11}{6} : \dfrac{44}{18} = \dfrac{3}{4}\)

Coi n_CO2 = 3 mol ; n_H2O = 4 mol

Ta có :

n_C = n_CO2 =3 mol

n_H = 2n_H2O = 4.2 = 8 mol

\(\Rightarrow \dfrac{n_C}{n_H} = \dfrac{3}{8}\)

Vậy CTHH của X:  C₃H₈O_x(x >0,x nguyên)

Chọn C.

Đáp án C

Bn check lại đề chứ mình nghĩ VCO2 = 0,896 (l)

\(n_{CO_2}=\dfrac{0,896}{22,4}=0,04\left(mol\right)\)

\(n_{H_2O}=\dfrac{0,72}{18}=0,04\left(mol\right)\)

Bảo toán C: n_C(X) = 0,04 (mol)

Bảo toàn H: n_H(X) = 0,04.2 = 0,08 (mol)

=> \(n_{O\left(X\right)}=\dfrac{0,88-0,04.12-0,08.1}{16}=0,02\left(mol\right)\)

=> n_C : n_H : n_O = 0,04 : 0,08 : 0,02 = 2 : 4 : 1

=> CTHH: (C₂H₄O)_n

Mà M = 44.2 = 88(g/mol)

=> n = 2

=> CTHH: C₄H₈O₂

C₄H₈O₂