\(n_{H_2}=\dfrac{0,4}{2}=0,2mol\)
\(n_{O_2}=\dfrac{1,12}{22,4}=0,05mol\)
\(2H_2+O_2\rightarrow\left(t^o\right)2H_2O\)
\(\dfrac{0,2}{2}\) > \(\dfrac{0,05}{1}\) ( mol )
=> H2 dư
\(n_{O_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\\ n_{H_2}=\dfrac{0,4}{2}=0,2\left(mol\right)\\ pthh:2H_2+O_2\underrightarrow{t^o}2H_2O\)
LTL : \(\dfrac{0,2}{2}>\dfrac{0,05}{1}\)
=> H2 dư