\(a) C_2H_4 + 3O_2 \xrightarrow{t^o} 2CO_2 + 2H_2O\\ V_{O_2} = 3V_{C_2H_4} = 4,48.3 = 13,44(lít)\\ b) V_{không\ khí} = \dfrac{V_{O_2}}{20\%} = 67,2(lít)\\ c) n_{C_2H_4} = \dfrac{4,48}{22,4} = 0,2(mol)\\ n_{H_2O} = 2n_{C_2H_4} = 0,4(mol)\\ m_{H_2O} = 0,4.18 = 7,2(gam) \Rightarrow V_{H_2O} = \dfrac{m}{d} = \dfrac{7,2}{1} = 7,2(ml) \)