\(BTNTC:n_C=n_{CO2}=\dfrac{6,72}{22,4}=0,3\rightarrow m_C=0,3.12=3,6< m_A\)

---> A chứa 2 nguyên tố C , H

\(m_H=m_A-m_C=4,8-3,6=1,2\left(g\right)\)

Đặt CT : \(C_xH_y\)

\(x:y=\dfrac{m_C}{12}=\dfrac{m_H}{1}=\dfrac{3,6}{12}:\dfrac{1,2}{1}=1:4\)

\(\rightarrow CTTQ:\left(CH_4\right)_n\)

\(\rightarrow12+4n=16\rightarrow n=1\)

\(\rightarrow CTPT:CH_4\)

\(m_{bình.tăng}=m_{C2H4}=\dfrac{4,48}{28}=0,16\left(mol\right)\)

a) Pt : \(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)

b) \(\left\{{}\begin{matrix}\%V_{C2H4}=\dfrac{0,16.24,79}{8,9244}.100\%=44,44\%\\\%V_{CH4}=100\%-44,44\%=55,56\%\end{matrix}\right.\)

c) Theo Pt : \(n_{C2H4}=n_{Br2}=0,16\left(mol\right)\)

\(\rightarrow V_{ddBr2}=\dfrac{0,16}{1,6}=0,1\left(l\right)\) 

\(m_{C6H5-CH3}=\dfrac{32,2}{70\%}=46\left(g\right)\)  sửa lại nhé

a) \(C_6H_5-CH_3+3HNO_3\xrightarrow[1:1]{H_2SO_4}C_6H_2\left(NO_2\right)_3CH_3+3H_2O\)

b) \(n_{C6H2\left(NO2\right)3CH3}=\dfrac{79,45}{227}=0,35\left(mol\right)\)

 \(C_6H_5-CH_3+3HNO_3\xrightarrow[H_2SO_4]{1:1}C_6H_2\left(NO_2\right)_3CH_3\)

Theo Pt : \(n_{C6H5-CH3}=n_{C6H2\left(NO2\right)3CH3}=0,35\left(mol\right)\)

\(\Rightarrow m_{C6H5-CH3\left(lt\right)}=0,35.92=32,2\left(g\right)\)

\(\Rightarrow m_{C6H5-CH3\left(tt\right)}=32,2.70\%=22,54\left(g\right)\)

a) \(n_{CO2}=\dfrac{6,72}{22,4}=0,3\left(mol\right);n_{H2O}=\dfrac{5,4}{18}=0,3\left(mol\right)\)

Ta thấy : \(n_{H2O}=n_{CO2}\rightarrow Anken\left(C_nH_{2n}\right)\)

\(\rightarrow n=\dfrac{n_{CO2}}{n_{anken}}=\dfrac{0,3}{0,1}=3\)

\(\rightarrow CTPT:C_3H_6\)

b) \(CTCT:CH_2=CH-CH_3\)