\(n_{H2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)
\(n_{O2}=\dfrac{6,4}{32}=0,2\left(mol\right)\)
PTHH : \(2H_2+O_2-t^o->2H_2O\)
Vì \(\dfrac{n_{H2}}{2}>n_{O2}\left(\dfrac{0,5}{2}>0,2\right)\) nên sau pứ H2 còn dư
Theo PTHH : \(n_{H2\left(pứ\right)}=2n_{O2}=0,4\left(mol\right)\)
\(\Rightarrow n_{H2\left(dư\right)}=0,5-0,4=0,1\left(mol\right)\)
\(\Rightarrow V_{H2}=0,1.22,4=2,24\left(l\right)\)
b) Theo PTHH : \(n_{H2O}=2n_{O2}=0,4\left(mol\right)\)
\(\Rightarrow m_{H2O}=0,4.18=7,2\left(g\right)\)
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