5. 

\(n_X=\dfrac{2,24}{22,4}=0,1mol\\ M_X=2,125.4=8,5g\cdot mol^{^{ }-1}\\ n_{H_2}=a;n_{C_2H_4}=b\\ a+b=0,1\\ 2a+28b=8,5.0,1=0,85\\ a=0,075;b=0,025\\ H_2+C_2H_4-^{^{ }Ni,t^{^{ }0}}->C_2H_6\\ V_{C_2H_6}=0,025.22,4=0,56L;V_{H_2dư}=22,4\left(0,075-0,025\right)=1,12L\)

6. 

Thu được Y chỉ gồm hydrocarbon nên khí hydrogen phản ứng hết.

\(n_A=\dfrac{4,48}{22,4}=0,2mol\\ n_Y=\dfrac{3,36}{22,4}=0,15mol\\ \Delta n_{hh}=n_{H_2\left(pư\right)}=0,05\left(mol\right)\\ n_{C_2H_4}=0,15\left(mol\right)\\ a.\%V_{H_2}=\dfrac{0,05}{0,2}=25\%\\ \%V_{C_2H_4}=75\%\\ b.BTLK\pi:0,15=0,05+n_{Br_2}\\ n_{Br_2}=0,1mol\)

\(C_2H_4 + Br_2 \to C_2H_4Br_2\\ n_{C_2H_4} = n_{Br_2} =\dfrac{24}{160}= 0,15(mol)\\ \%V_{C_2H_4} = \dfrac{0,15.22,4}{6,72}.100\% = 50\%\\ \%V_{CH_4}= 100\% -50\% = 50\%\)

\(n_{hh}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)

\(n_{Br_2}=\dfrac{8}{160}=0,05\left(mol\right)\)

\(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)

0,05      0,05                   ( mol )

\(\left\{{}\begin{matrix}\%V_{C_2H_4}=\dfrac{0,05}{0,5}.100=10\%\\\%V_{CH_4}=100\%-10\%=90\%\end{matrix}\right.\)

--> Chọn B

Chọn B

1) \(\left\{{}\begin{matrix}\%V_{CH_4}=\dfrac{1,68}{6,72}\cdot100\%=25\%\\\%V_{C_2H_2}=75\%\end{matrix}\right.\)

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

\(\Rightarrow\%V_{C_2H_4}=\dfrac{0,2}{\dfrac{5,6}{22,4}}\cdot100\%=80\%\) \(\Rightarrow\%V_{CH_{_4}}=20\%\)

huhuu bạn nào giúp tớ với ....

Ta có: \(n_Y=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

 \(\Rightarrow M_Y=\dfrac{1,32}{0,1}=13,2\left(g/mol\right)\)

→ Y gồm H₂ dư và C₂H₆.

\(\Rightarrow\left\{{}\begin{matrix}n_{H_2\left(dư\right)}+n_{C_2H_6}=0,1\\2n_{H_2\left(dư\right)}+30n_{C_2H_6}=1,32\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}n_{H_2\left(dư\right)}=0,06\left(mol\right)\\n_{C_2H_6}=0,04\left(mol\right)\end{matrix}\right.\)

Gọi: \(\left\{{}\begin{matrix}n_{H_2}=x\left(mol\right)\\n_{C_2H_2}=n_{C_2H_4}=y\left(mol\right)\end{matrix}\right.\)

BTNT C: 2y = 0,04 ⇒ y = 0,02 (mol)

Có: n_{H2 (pư)} = 2n_C2H2 + n_C2H4 = 0,06 (mol)

⇒ n_H2 = 0,06 + 0,06 = 0,12 (mol)

\(\Rightarrow V=\left(0,12+0,02+0,02\right).22,4=3,584\left(l\right)\)

\(PTHH:C_2H_4+Br_2\rightarrow C_2H_4Br_2\\ n_{CH_4}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\\ \%m_{CH_4}=\dfrac{0,1.16}{3}.100\%=53,33\%\\ \%m_{C_2H_4}=100\%-53,33\%=46,67\%\)

\(V_{khí.thoát.ra}=V_{CH_4}=2,24l\)

\(n_{CH_4}=\dfrac{2,24}{22,4}=0,1mol\)

\(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)

\(m_{CH_4}=0,1.16=1,6g\)

\(\left\{{}\begin{matrix}\%m_{CH_4}=\dfrac{1,6}{3}.100=53,33\%\\\%m_{C_2H_4}=100\%-53,33\%=46,67\%\end{matrix}\right.\)