a, PT: \(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)

Khí thoát ra khỏi bình là CH₄ (metan).

b, Ta có: \(\left\{{}\begin{matrix}\%V_{CH_4}=\dfrac{1,12}{5,6}.100\%=20\%\\\%V_{C_2H_4}=100-20=80\%\end{matrix}\right.\)

c, Ta có: \(V_{C_2H_4}=5,6.80\%=4,48\left(l\right)\)

\(\Rightarrow n_{C_2H_4}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

Theo PT: \(n_{Br_2}=n_{C_2H_4}=0,2\left(mol\right)\)

\(\Rightarrow m_{Br_2}=0,2.160=32\left(g\right)\)

Bạn tham khảo nhé!

a) khí thoát ra khỏi bình là khí metan (CH4)

VC2H2 = 1.12-0.56=0.56 (l)

%VCH4 = 0,56 .100/1,12 = 50%

%VC2H2= 0,56 . 100/1,12 = 50%

b) nC2H2 = 0,56/22,4 =0,025 (mol)

C2H2 + 2Br2 -> C2H2Br4

0,025 ----> 0.05

=> mBr2 = 0,05 . 160= 8 (g)

chất khí thoát ra là metan đó bạn sau đó bạn tíh số mol của metan => etylen

C2H4 + Br2 = C2H4Br2

a.\(n_{hh}=\dfrac{13,44}{22,4}=0,6mol\)

\(n_{C_2H_2Br_4}=\dfrac{6,72}{22,4}=0,3mol\)
\(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)

 0,3                         0,3      ( mol )

\(\%C_2H_2=\dfrac{0,3}{0,6}.100=50\%\)

\(\%CH_4=100\%-50\%=50\%\)

b.

\(CH_4+2O_2\rightarrow\left(t^o\right)CO_2+2H_2O\)

 0,3       0,6                                       ( mol )

\(2C_2H_2+5O_2\rightarrow\left(t^o\right)4CO_2+2H_2O\)

 0,3          0,75                                        ( mol )

\(V_{O_2}=\left(0,6+0,75\right).22,4=1,35.22,4=30,24l\)

Khí thoát ra là metan

\(\%V_{CH_4} = \dfrac{5,6}{13,44}.100\% =41,67\%\\ \%V_{C_2H_4} = 100\% - 41,67\% = 58,33\%\\ b) V_{C_2H_2} = 13,44 -5,6 = 7,84(lít)\\ n_{C_2H_2} = \dfrac{7,84}{22,4} = 0,35(mol)\\ C_2H_2 + 2Br_2 \to C_2H_2Br_4\\ n_{Br_2} = 2n_{C_2H_2} = 0,7(mol)\\ \Rightarrow C_{M_{Br_2}} = \dfrac{0,7}{0,25} = 2,8M\\ c)\)

\(CH_4 + 2O_2 \xrightarrow{t^o} CO_2 + 2H_2O\\ C_2H_4 + 3O_2 \xrightarrow{t^o} 2CO_2 + 2H_2O\\ V_{O_2} = 2V_{CH_4} + 3V_{C_2H_4} = 5,6.2 + 7,84.3 = 34,72(lít)\)

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

a, PT: \(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)

b, - Khí thoát ra là CH₄.

⇒ V_CH4 = 4,48 (l)

\(\Rightarrow\left\{{}\begin{matrix}\%V_{CH_4}=\dfrac{4,48}{11,2}.100\%=40\%\\\%V_{C_2H_4}=100-40=60\%\end{matrix}\right.\)

a)

Khí thoát ra: CH₄

\(\%V_{CH_4} = \dfrac{6,72}{16,8}.100\% = 40\%\\ \%V_{C_2H_4} = 100\% - 40\% = 60\%\)

b)

\(n_{C_2H_4} = \dfrac{16,8-6,72}{22,4} = 0,45(mol)\\ C_2H_4 + Br_2 \to C_2H_4Br_2\\ n_{Br_2} = n_{C_2H_4} = 0,45(mol)\\ \Rightarrow C_{M_{Br_2}} = \dfrac{0,45}{2} = 0,225M\\ c) n_{C_2H_4Br_2} = n_{C_2H_4} = 0,45(mol)\\ \Rightarrow n_{C_2H_4Br_2} = 0,45.188 = 84,6(gam)\)

Bài 4:

a) n(hỗn hợp khí)= 16,8/22,4=0,75(mol)

- Khí thoát ra là khí CH4.

=> nCH4=6,72/22,4=0,3(mol)

nC2H4=0,75-0,3=0,45(mol)

- Số mol tỉ lệ thuận với thể tích.

%V(CH4)=%nCH4= (0,3/0,75).100=40%

=> %V(C2H4)=100% - 40%=60%

b) PTHH: C2H4 + Br2 -> C2H4Br2

nC2H4Br2= nBr2=nC2H4=0,45(mol)

=>VddBr2= 0,45/2=0,225(l)

c) mC2H4Br2=0,45. 188= 84,6(g)

Khí thoát ra ngoài : $CH_4$

$C_2H_2 + 2Br_2 \to C_2H_2Br_4$

$V_{CH_4} = 1,12(lít) \Rightarrow \%V_{CH_4} = \dfrac{1,12}{3,36}.100\% = 33,33\%$
$\%V_{C_2H_2} = 100\% -33,33\% = 66,67\%$