a) \(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)
1 2 1 (mol)
0,1 0,2 0,1 (mol)
ta có \(n_{C_2H_2Br_4}=\dfrac{43,5}{346}\approx0,1\left(mol\right)\)
=> \(m_{Br_2}\approx0,2.160=32\left(g\right)\)
Ta có ;
\(V_{C_2H_2}\approx0,1.22,4=2,24\left(l\right)\)
=> V\(\%_{C_2H_2}=\dfrac{2,24}{6,5}.100\%\approx34,5\%\)
=> V\(\%_{CH_4}\approx100\%-34,5\%=65,5\%\)