\(n_{hh\left(khi\right)}=\frac{11,2}{22,4}=0,5\left(mol\right)\)
Gọi \(\left\{{}\begin{matrix}n_{C2H4}:x\left(mol\right)\\n_{C2H2}:y\left(mol\right)\end{matrix}\right.\left(x,y>0\right)\)
\(C_2H_4+Br_2\rightarrow C_2H_4Br_2\left(1\right)\)
x________x____________
\(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\left(2\right)\)
y________2y___________
Khối lượng bình tăng là \(m_{C2H4}+m_{C2H2}=4g\left(I\right)\)
Khí thoát ra khỏi bình là CH4 \(\left(V_{CH4}=7,84\left(l\right)\right)\)
\(\Rightarrow V_{C2H4}+V_{C2H2}=11,2-7,84=3,36\left(l\right)\)
\(n_{C2H4}+n_{C2H2}=\frac{3,36}{22,4}=0,15\left(mol\right)\left(II\right)\)
Giải hệ PT:
\(\left\{{}\begin{matrix}x+y=0,15\\28x+26y=4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,05\\y=0,1\end{matrix}\right.\)
\(V_{C2H4}=0,05.22,4=1,12\left(l\right)\)
\(V_{C2H2}=0,1.22,4=2,24\left(l\right)\)
