Ta có:
\(\left\{{}\begin{matrix}n_{khi}=\frac{3,36}{22,4}=0,15\left(mol\right)\\n_{Br2}=\frac{8}{160}=0,05\left(mol\right)\end{matrix}\right.\)
\(PTHH:C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
_________0,05____0,05__________
\(\Rightarrow\left\{{}\begin{matrix}\%V_{C2H4}=\%n_{C2H4}=\frac{0,05}{0,15}.100\%=33,33\%\\\%V_{CH4}=100\%-33,33\%=66,67\%\end{matrix}\right.\)
\(n_{CH4}=0,15-0,05=0,1\left(mol\right)\)
\(CH_4+2O_2\rightarrow CO_2+2H_2O\)
0,1____0,2_____________
\(C_2H_4+3O_2\rightarrow2CO_2+2H_2O\)
0,05____0,15____________
\(V_{O2}=\left(0,2+0,15\right).22,4=7,84\left(l\right)\)
\(\Rightarrow V_{kk}=\frac{7,84}{21\%}==37,33\left(l\right)\)
