\(a.n_{Br_2}=0,1.2=0,2mol\\ C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
0,2 0,2 0,2
\(V_{C_2H_4}=0,2.22,4=4,48l\\ V_{CH_4}=11,2-22,4=6,72l\\ b.\%V_{C_2H_4}=\dfrac{4,48}{11,2}\cdot100\%=40\%\\ \%V_{CH_4}=60\%\\ c.n_{CH_4}=\dfrac{6,72}{22,4}=0,3mol\\ C_2H_4+3O_2\xrightarrow[]{t^0}2CO_2+2H_2O\\ CH_4+2O_2\xrightarrow[]{t^0}CO_2+2H_2O\\ V_{O_2}=\left(0,2.3+0,3.2\right).22,4=26,88l\)