\(n_{H_2}=\dfrac{12,395}{24,79}=0,5\left(mol\right)\)
\(n_{O_2}=\dfrac{9,916}{24,79}=0,4\left(mol\right)\)
PT: \(2H_2+O_2\underrightarrow{t^o}2H_2O\)
Xét tỉ lệ: \(\dfrac{0,5}{2}< \dfrac{0,4}{1}\), ta được O2 dư.
Theo PT: \(n_{H_2O\left(LT\right)}=n_{H_2}=0,5\left(mol\right)\)
\(\Rightarrow H=\dfrac{0,4}{0,5}.100\%=80\%\)