\(n_{C_2H_4}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\\ PTHH:C_2H_4+2O_2\rightarrow^{t^o}CO_2+2H_2O\\ a,n_{O_2}=2n_{C_2H_4}=0,6\left(mol\right)\\ \Rightarrow V_{O_2\left(đktc\right)}=0,6\cdot22,4=13,44\left(l\right)\\ b,\left\{{}\begin{matrix}n_{CO_2}=n_{C_2H_4}=0,3\left(mol\right)\\n_{H_2O}=n_{O_2}=0,6\left(mol\right)\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}V_{CO_2\left(đktc\right)}=0,3\cdot22,4=6,72\left(l\right)\\m_{H_2O}=0,6\cdot18=10,8\left(g\right)\end{matrix}\right.\)
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