\(n_{CH_4}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)
\(n_{hh\left(CH_4,C_2H_2,C_2H_4\right)}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\ \rightarrow n_{hh\left(C_2H_2,C_2H_4\right)}=0,2-0,05=0,15\left(mol\right)\left(1\right)\)
\(m_{tăng}=m_{C_2H_2}+m_{C_2H_4}=4,1\left(g\right)\left(2\right)\)
\(\left(1\right)\left(2\right)\rightarrow n_{C_2H_2}=0,05\left(mol\right);n_{C_2H_4}=0,1\left(mol\right)\)
PTHH:
C2H2 + 2Br2 ---> C2H2Br4
C2H4 + Br2 ---> C2H4Br2
\(\left\{{}\begin{matrix}m_{CH_4}=0,05.16=0,8\left(g\right)\\m_{C_2H_2}=0,02.26=1,3\left(g\right)\\m_{C_2H_4}=0,1.28=2,8\left(g\right)\end{matrix}\right.\\ \rightarrow\left\{{}\begin{matrix}\%m_{CH_4}=\dfrac{0,8}{4,1+0,8}.100\%=16,33\%\\\%m_{C_2H_2}=\dfrac{1,3}{4,1+0,8}.100\%=26,53\%\\\%m_{C_2H_4}=100\%-16,33\%-26,52\%=57,15\%\end{matrix}\right.\)
