a) \(n_{Br_2}=\dfrac{56}{160}=0,35\left(mol\right)\)

Chỉ có \(C_2H_4\) phản ứng với \(brom\)

\(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)

b) \(V_{C_2H_4}=0,35.2,24=7,84\left(l\right)\Rightarrow\%C_2H_4=70\%\)

\(\%CH_4=100\%-70\%=30\%\)

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

Ta có: \(n_{Br_2}=\dfrac{16}{160}=0,1\left(mol\right)=n_{C_2H_4}\)

\(\Rightarrow\%V_{C_2H_4}=\dfrac{0,1\cdot22,4}{8,96}=25\%\) \(\Rightarrow\%V_{CH_4}=75\%\)

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

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

\(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)

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

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

\(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)

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

\(\Rightarrow x+y=\dfrac{1,344}{22,4}=0,06\left(1\right)\)

Ta có: \(n_{Br_2}=\dfrac{16}{160}=0,1\left(mol\right)\)

Theo PT: \(n_{Br_2}=n_{C_2H_4}+2n_{C_2H_2}=x+2y\left(mol\right)\)

⇒ x + 2y = 0,1 (2)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,02\left(mol\right)\\y=0,04\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%V_{C_2H_4}=\dfrac{0,02}{0,06}.100\%\approx33,33\%\\\%\text{ }V_{C_2H_2}\approx66,67\%\end{matrix}\right.\)

b, Ta có: 1/2 hỗn hợp khí gồm: 0,01 mol C₂H₄ và 0,02 mol C₂H₂.

PT: \(C_2H_4+3O_2\underrightarrow{t^o}2CO_2+2H_2O\)

\(2C_2H_2+5O_2\underrightarrow{t^o}4CO_2+2H_2O\)

Theo PT: \(n_{CO_2}=2n_{C_2H_4}+2n_{C_2H_2}=0,06\left(mol\right)\)

\(CO_2+Ca\left(OH\right)_2\rightarrow CaCO_{3\downarrow}+H_2O\)

Theo PT: \(n_{CaCO_3}=n_{CO_2}=0,06\left(mol\right)\)

\(\Rightarrow m_{cr}=m_{CaCO_3}=0,06.100=6\left(g\right)\)

Bạn tham khảo nhé!

 Gọi số mol của C2H4 và C2H2 lần lượt là x và y mol
theo bài ra: x+y = 0,56/22,4 = 0,025 (mol) 
Pt:
C2H4 + Br2 → C2H4Br2
x mol x mol x mol
C2H2 + 2 Br2 → C2H2Br4
y mol 2y mol y mol
Số mol n Br2 = x+2y = 5,6/160 = 0,035 9mol) 
Giải hệ ta đc: x = 0,015 và y = 0,01
=> %V C2H4 = 0,015/0,025 = 60% ; %V C2H2 = 40%

8g brom sau phản ứng :) ?

TK

\(\left\{{}\begin{matrix}C_2H_4:x\left(mol\right)\\C_2H_2:y\left(mol\right)\end{matrix}\right.\)⇒ x + y = \(\dfrac{6,72}{22,4}=0,3\left(1\right)\)

\(C_2H_4 + Br_2 \to C_2H_4Br_2\\ C_2H_2 + 2Br_2 \to C_2H_2Br_4\)

Theo PTHH : 

x + 2y = \(\dfrac{64}{160} = 0,4(2)\)

Từ (1)(2) suy ra: x = 0,2 ; y = 0,1

Vậy :

\(\%V_{C_2H_4} = \dfrac{0,2}{0,3}.100\% = 66,67\%\\ \%V_{C_2H_2} = 100\% - 66,67\% = 33,33\%\)

\(n_{CO_2}=0.3\left(mol\right)\)

\(Đặt:n_{C_2H_2}=a\left(mol\right),n_{C_2H_4}=b\left(mol\right)\)

\(n_{Br_2}=\dfrac{64}{160}=0.4\left(mol\right)\)

\(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)

\(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)

\(\left\{{}\begin{matrix}a+b=0.3\\2a+b=0.4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=0.1\\b=0.2\end{matrix}\right.\)

\(\%V_{C_2H_2}=\dfrac{0.1}{0.3}\cdot100\%=33.33\%\)

\(\%V_{C_2H_4}=66.67\%\)

Ta có: \(n_{Br_2}=\dfrac{48}{160}=0,3\left(mol\right)\)

PT: \(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)

____0,15___0,3 (mol)

\(\Rightarrow\left\{{}\begin{matrix}V_{C_2H_2}=0,15.22,4=3,36\left(l\right)\\V_{CH_4}=2,24\left(l\right)\end{matrix}\right.\)

Bạn tham khảo nhé!

$C_2H_2+2Br_2\to Br_2CH-CHBr_2$

$n_{Br_2}=\dfrac{48}{160}=0,3(mol)$

$\Rightarrow n_{C_2H_2}=\dfrac{n_{Br_2}}{2}=0,15(mol)$

$\Rightarrow V_{C_2H_2}=0,15.22,4=3,36(l)$

$\Rightarrow V_{CH_4}=5,6-3,36=2,24(l)$

C2,H2 hay C2H2?