\(n_{H_2}=\frac{2,24}{22,4}=0,1\left(mol\right)\)
\(n_{O_2}=\frac{3,2}{32}=0,1\left(mol\right)\)
PTHH : \(2H_2+O_2\underrightarrow{t^o}2H_2O\)
Xét tỉ lệ
\(\left\{{}\begin{matrix}\frac{n_{H_2}}{2}=\frac{0,1}{2}=0,05\\\frac{n_{O_2}}{1}=\frac{0,1}{1}=0,1\end{matrix}\right.\)\(\rightarrow\)\(\frac{n_{H_2}}{2}< \frac{n_{O_2}}{1}\)
=> H2 phản ứng hết,O2 còn dư
PTHH : \(2H_2+O_2\underrightarrow{t^o}2H_2O\)
Theo pt 2 1 2 :mol
Theo đb 0,1 0,1 :mol
Phản ứng \(0,1\) \(\)\(\frac{0,1\cdot1}{2}\) \(\frac{0,1\cdot2}{2}\) :mol
Sau pứ 0 0,05 0,1 :mol
\(V_{O_{2_{dư}}}=\left(0,1-0,05\right)\cdot22,4=1,12\left(l\right)\)