m_tăng = m_C2H4

=> \(n_{C_2H_4}=\dfrac{5,6}{28}=0,2\left(mol\right)\)

=> V_C2H4 = 0,2.22,4 = 4,48 (l)

=> V_CH4 = 6,72 - 4,48 = 2,24 (l)

\(n_{CH_4}=\dfrac{5.6}{22.4}=0.25\left(mol\right)\)

\(m_{tăng}=m_{C_2H_4}=7\left(g\right)\)

\(m_{hh}=m_{CH_4}+m_{C_2H_4}=0.25\cdot16+7=11\left(g\right)\)

C2H4: 1,7% và CH4:88,3% 

a) nC2H4Br2=47/188=0,25(mol)

n(CH4,C2H4)=11,2/22,4=0,5(mol)

PTHH: C2H4 + Br2 -> C2H4Br2

0,25<----------0,25<---------0,25(mol)

mBr2(p.ứ)=0,25 x 160= 40(g)

b) V(C2H4,đktc)=0,25 x 22,4= 5,6(l)

=> %V(C2H4)=(5,6/11,2).100=50%

=>%V(CH4)=100% - 50%= 50%

Gọi: \(\left\{{}\begin{matrix}n_{CH_4}=x\left(mol\right)\\n_{C_2H_4}=y\left(mol\right)\\n_{C_2H_2}=z\left(mol\right)\end{matrix}\right.\) \(\Rightarrow x+y+z=\dfrac{8,4}{22,4}=0,375\left(mol\right)\left(1\right)\)

Ta có: m _{bình Br2 tăng} = m_C2H4 + m_C2H2

⇒ 8,1 = 28y + 26z (2)

- Khí thoát ra khỏi bình là CH₄.

PT: \(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)

\(n_{CO_2}=\dfrac{1,68}{22,4}=0,075\left(mol\right)=n_{CH_4}=x\left(3\right)\)

Từ (1), (2) và (3) \(\Rightarrow\left\{{}\begin{matrix}x=0,075\left(mol\right)\\y=0,15\left(mol\right)\\z=0,15\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow m_X=m_{CH_4}+m_{C_2H_4}+m_{C_2H_2}=0,075.16+8,1=9,3\left(g\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{CH_4}=\dfrac{0,075.16}{9,3}.100\%\approx12,9\%\\\%m_{C_2H_4}=\dfrac{0,15.28}{9,3}.100\%\approx45,16\%\\\%m_{C_2H_2}\approx41,94\%\end{matrix}\right.\)

\(n_{CH_4}=a;n_{C_2H_4}=b;n_{C_2H_2}=c\\ a+b+c=\dfrac{8,4}{22,4}\left(1\right)\\ C_2H_4+Br_2->C_2H_4Br_2\\ C_2H_2+2Br_2->C_2H_2Br_4\\ m_{bình.tăng}=28b+26c=8,1g\\ n_{CO_2}=a=\dfrac{1,68}{22,4}\\ a=0,075;b=c=0,15\\ \%V_{CH_4}=\dfrac{0,075}{0,375}.100\%=20\%\\ \%V_{C_2H_4}=\%V_{C_2H_2}=\dfrac{0,15}{0,375}.100\%=40\%\)

Ai giúp e với ạ

\(n_{hhkhí\left(C_2H_4,CH_4\right)}=\dfrac{7,84}{22,4}=0,35\left(mol\right)\)

\(m_{tăng}=m_{C_2H_4}=4,2\left(g\right)\\ n_{C_2H_4}=\dfrac{4,2}{28}=0,15\left(mol\right)\\ \%V_{C_2H_4}=\dfrac{0,15}{0,35}=42,85\%\\ \%V_{CH_4}=100\%-42,85\%=57,15\%\)

\(m_{bìnhtăng}=m_{anken}=m_{etilen}=1,4g\)

\(\Rightarrow n_{C_2H_4}=\dfrac{1,4}{28}=0,05mol\)

\(n_{hh}=\dfrac{4,48}{22,4}=0,2mol\)

\(\Rightarrow n_{metan}=n_{hh}-n_{eilen}=0,2-0,05=0,15mol\)

\(\%V_{metan}=\dfrac{0,15}{0,2}\cdot100\%=75\%\)

\(\%V_{etilen}=100\%-75\%=25\%\)

PTHH: \(CH_4+2O_2\xrightarrow[]{t^o}CO_2+2H_2O\)

Ta có: \(n_{CO_2}=\dfrac{1,68}{22,4}=0,075\left(mol\right)=n_{CH_4}\)

Đặt \(\left\{{}\begin{matrix}n_{C_2H_4}=a\left(mol\right)\\n_{C_2H_2}=b\left(mol\right)\end{matrix}\right.\) \(\Rightarrow a+b=\dfrac{5,04}{22,4}-0,075=0,15\)  (1)

PTHH: \(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)

            \(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)

Theo PTHH: \(28a+26b=4,1\)  (2)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}a=n_{C_2H_4}=0,1\left(mol\right)\\b=n_{C_2H_2}=0,05\left(mol\right)\end{matrix}\right.\)

Mặt khác: \(n_{hh}=\dfrac{5,04}{22,4}=0,225\left(mol\right)\) \(\Rightarrow\left\{{}\begin{matrix}\%V_{CH_4}=\dfrac{0,075}{0,225}\cdot100\%\approx33,33\%\\\%V_{C_2H_4}=\dfrac{0,1}{0,225}\cdot100\%\approx44,44\%\\\%V_{C_2H_2}=22,23\%\end{matrix}\right.\)