2H2+O2-to-->2H2O
a)nH2=0,4(mol)
nO2=0,3(mol)
=> O2 dư(do tỉ số)
Theo pthh
nH2O=nH2=0,4(mol)
mH2O=0,4.18=7,2(g)
\(2H_2+O_2\rightarrow2H_2O\)
\(nH_2=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)
\(nO_2=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
\(\dfrac{0,4}{2}< \dfrac{0,3}{1}\) => Oxi dư, hydro đủ
\(\Rightarrow nH_2O=nH_2=0,4\left(mol\right)\)
\(\Rightarrow nH_2O=0,4.18=7,2\left(g\right)\)