\(n_{hhkhí}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\\ m_{tăng}=m_{C_2H_2}=2,8\left(g\right)\\ \rightarrow n_{C_2H_2}=\dfrac{2,8}{28}=0,1\left(mol\right)\\ PTHH:C_2H_2+2Br_2\rightarrow C_2H_2Br_4\\ n_{CH_4}=0,15-0,1=0,05\left(mol\right)\\ \rightarrow\left\{{}\begin{matrix}\%V_{C_2H_2}=\dfrac{0,1}{0,15}=66,67\%\\\%V_{CH_4}=\dfrac{0,05}{0,15}=33,33\%\end{matrix}\right.\)
\(PTHH:\\ 2C_2H_2+5O_2\underrightarrow{t^o}4CO_2+2H_2O\\ 0,1\rightarrow0,25\\ CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\\ 0,05\rightarrow0,1\)
=> VO2 = (0,1 + 0,25).22,4 = 7,84 (l)
\(n_{hh}=\dfrac{3,36}{22,4}=0,15mol\)
\(n_{tăng}=m_{C_2H_4}=2,8g\)
\(n_{C_2H_4}=\dfrac{2,8}{28}=0,1mol\)
\(\%V_{C_2H_4}=\dfrac{0,1}{0,15}=66,67\%\)
\(\%V_{CH_4}=100\%-66,67\%=33,33\%\)
