\(m_{CH_4} = 8.30\% = 2,4(gam)\\ m_{C_2H_4} = 8 - 2,4 = 5,6(gam)\\ CH_4 + 2O_2 \xrightarrow{t^o} CO_2 + 2H_2O\\ C_2H_4 + 3O_2 \xrightarrow{t^o} 2CO_2 + 2H_2O\\\)
Theo PTHH :
\(n_{CO_2} = n_{CH_4} + 2n_{C_2H_4} = \dfrac{2,4}{16} + \dfrac{5,6}{28}.2 = 0,55(mol)\\ \Rightarrow m_{CO_2} = 0,55.44 = 24,2(gam)\)
\(m_{CH_4}=0.3\cdot8=2.4\left(g\right)\)
\(n_{CH_4}=\dfrac{2.4}{16}=0.15\left(mol\right)\)
\(m_{C_2H_4}=8-2.4=5.6\left(g\right)\)
\(n_{C_2H_4}=\dfrac{5.6}{28}=0.2\left(mol\right)\)
\(CH_4+2O_2\underrightarrow{t^0}CO_2+2H_2O\)
\(C_2H_4+3O_2\underrightarrow{t^0}2CO_2+2H_2O\)
\(n_{CO_2}=2\cdot0.15+0.2=0.5\left(mol\right)\)
\(m_{CO_2}=0.5\cdot44=22\left(g\right)\)
