\(a.2H_2+O_2-^{t^o}\rightarrow2H_2O\\ b.n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\\ n_{O_2}=\dfrac{1}{2}n_{H_2}=0,2\left(mol\right)\\ \Rightarrow V_{O_2}=0,2.22,4=4,48\left(l\right)\\ TrongkhôngkhíO_2chiếm20\%\\ \Rightarrow V_{kk}=\dfrac{4,48}{20\%}=22,4\left(l\right)\)