\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\
pthh:Mg+2HCl\rightarrow MgCl_2+H_2\)
0,2 0,2
\(n_{O_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\\ pthh:2H_2+O_2\underrightarrow{t^o}2H_2O\)
\(LTL:\dfrac{0,2}{2}< \dfrac{0,15}{1}\)
=> Oxi dư
\(n_{H_2O}=n_{H_2}=0,2\left(mol\right)\\
m_{H_2O}=0,2.18=3,6g\)
đề hơi sai sai bạn ạ :))