Ta có:
\(\left\{{}\begin{matrix}n_A=\frac{6,72}{22,4}=0,3\left(mol\right)\\n_{CH4}=\frac{2,24}{22,4}=0,1\left(mol\right)\end{matrix}\right.\)
Gọi \(\left\{{}\begin{matrix}n_{C2H2}:x\left(mol\right)\\n_{C2H4}:y\left(mol\right)\end{matrix}\right.\)
\(CH\equiv CH+2Br_2\rightarrow CHBr_2+CHBr_2\)
x________2x___________________
\(CH_2=CH_2+Br_2\rightarrow CH_2Br+CH_2Br\)
y_____________y______________
Đổi 200ml=0,2l
\(n_{Br2}=0,2.1,5=0,3\left(mol\right)\)
\(X+Y=0,3-0,1\)
\(2X+Y=0,3\)
\(\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,1\end{matrix}\right.\)
\(\%V_{CH4}=\%V_{C2H2}=\%V_{C2H4}=\frac{0,1}{0,3}.100\%=\frac{100}{3}\%\)
\(n_{CO2}=n_{CH4}+2n_{C2H2}+2n_{C2H4}=0,5\)
\(n_{H2O}=4n_{CH4}+2n_{C2H2}+4n_{C2H4}=1\)
\(2n_{O2}=2n_{CO2}+n_{H2O}\Leftrightarrow n_{O2}=1\)
\(V_{O2}=1.22,4=22,4\left(l\right)\)
