a, \(2H_2+O_2\underrightarrow{^{to}}2H_2O\)
b,\(n_{H2}=\frac{11,2}{22,4}=0,5\left(mol\right)\)
\(n_{O2}=\frac{11,2}{22,4}=0,5\left(mol\right)\)
Lập tỉ lệ : \(\frac{n_{H2}}{2}< \frac{n_{O2}}{1}\Rightarrow\) O2 dư , H2 hết
\(\Rightarrow n_{O2\left(Dư\right)}=n_{O2}-\frac{1}{2}n_{H2}=0,5-0,25=0,25\left(mol\right)\)
\(\Rightarrow V_{O2}=0,25.22,5=5,6\left(l\right)\)
c,\(n_{H2O}=n_{H2}=0,5\left(mol\right)\)
\(\Rightarrow m_{H2O}=0,5.18=9\left(g\right)\)