Question

Answer 1

9. \(\left\{{}\begin{matrix}n_C=n_{CO_2}=\dfrac{9,9}{44}=0,225\left(mol\right)\\n_H=2n_{H_2O}=2.\dfrac{5,4}{18}=0,6\left(mol\right)\\n_{O\left(A\right)}=\dfrac{4,5-0,225.12-0,6}{16}=0,075\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow n_C:n_H:n_O=0,225:0,6:0,075=3:8:1\)

\(\Rightarrow A:\left(C_3H_8O\right)_n\)

\(\Rightarrow n=\dfrac{60}{60}=1\)

Vậy A là C₃H₈O

10) \(\left\{{}\begin{matrix}n_C=n_{CO_2}=\dfrac{22}{44}=0,5\left(mol\right)\\n_H=2n_{H_2O}=2.\dfrac{13,5}{18}=1,5\left(mol\right)\\n_O=\dfrac{7,5-0,5.12-1,5}{16}=0\left(mol\right)\end{matrix}\right.\)

``=> A`` không chứa O

\(\Rightarrow n_C:n_H=0,5:1,5=1:3\)

\(\Rightarrow A:\left(CH_3\right)_n\)

\(\Rightarrow n=\dfrac{15.2}{15}=2\)

Vậy A là C₂H₆

11) \(\left\{{}\begin{matrix}n_C=n_{CO_2}=\dfrac{0,224}{22,4}=0,01\left(mol\right)\\n_H=2n_{H_2O}=2.\dfrac{0,18}{18}=0,02\left(mol\right)\\n_O=\dfrac{0,3-0,01.12-0,02}{16}=0,01\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow n_C:n_H:n_O=0,01:0,02:0,01=1:2:1\)

\(\Rightarrow A:\left(CH_2O\right)_n\)

\(\Rightarrow n=\dfrac{30.2}{30}=2\)

Vậy A là C₂H₄O₂

12) Gọi \(\left\{{}\begin{matrix}n_{CO_2}=4a\left(mol\right)\\n_{H_2O}=5a\left(mol\right)\end{matrix}\right.\)

Theo ĐLBTKL: \(44.4a+18.5a=2,25+32.\dfrac{3,08}{22,4}\Leftrightarrow a=0,025\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_C=n_{CO_2}=0,025.4=0,1\left(mol\right)\\n_H=2n_{H_2O}=5.2.0,025=0,25\left(mol\right)\\n_O=\dfrac{2,25-0,1.12-0,25}{16}=0,05\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow n_C:n_H:n_O=0,1:0,25:0,05=2:5:1\)

\(\Rightarrow A:\left(C_2H_5O\right)_n\)

\(\Rightarrow n=\dfrac{45.2}{45}=2\)

Vậy A là C₄H₁₀O₂

13) Đề sai