Giải thích các bước giải: 

Câu 1:

\(\begin{array}{l} \\CTTQ\ Ankan:\ C_nH_{2n+2}\\M_X=2×29=58=14n+2(g/mol)\\\to n=4\\\to Ankan:\ C_4H_{10}\end{array}\)

Câu 2:

\(\begin{array}{l} n_{H_2O}=\dfrac{10,8}{18}=0,6(mol)\\n_{CO_2}=\dfrac{11,2}{22,4}=0,5\\n_{H_2O}>n_{CO_2}\to X:\ Ankan\\n_X=n_{H_2O}-n_{CO_2}=0,1(mol)\\C=\dfrac{n_{CO_2}}{n_{X}}=\dfrac{0,5}{0,1}=5\\\to X:\ C_5H_{12}\\ CTPT:\ (CH_3)_3C-CH_3\end{array}\)

Câu 3:

\(\begin{array}{l} n_{CO_2}=\dfrac{57,2}{44}=1,3(mol)\\n_{H_2O}=\dfrac{30,6}{18}=1,7(mol)\\n_{X}=n_{H_2O}-n_{CO_2}=0,4(mol)\\\overline{C}=\dfrac{1,3}{0,4}=3,25\\\to 2\ Ankan:\ C_3H_8,\ C_4H_{10}\end{array}\)

Đáp án + Giải thích các bước giải:

Bài 1:

\(\begin{array} \\ X:C_nH_{2n+2}\\ \to M_X=14n+2=2.29=58(g/mol)\\ \to n=4\\ \to X:C4H_{10} \end{array}\)

Bài 2:

\(\begin{array} \\ n_{CO_2}=\dfrac{11,2}{22,4}=0,5(mol)\\ n_{H_2O}=\dfrac{10,8}{18}=0,6(mol)\\ \to X:Ankan\\ C=\dfrac{0,5}{0,6-0,5}=5\\\to CTPT\;X:C_5H_{12}\\ CTCT:C(CH_3)_4\end{array}\)

Bài 3:

\(\begin{array} \\ n_{CO_2}=\dfrac{57,2}{44}=1,3(mol)\\ n_{H_2O}=\dfrac{30,6}{18}=1,7(mol)\\ \to X:2\,ankan\\ \overline{C}=\dfrac{1,3}{1,7-1,3}=3,25\\ \to X:C_3H_8,C_4H_{10} \end{array}\)