PTHH: \(Mg+2HCl\rightarrow MgCl_2+H_2\) (1)
\(MgO+2HCl\rightarrow MgCl_2+H_2O\) (2)
a) Ta có: \(n_{H_2}=\frac{2,24}{22,4}=0,1\left(mol\right)\) \(\Rightarrow n_{Mg}=0,1mol\)
\(\Rightarrow m_{Mg}=0,1\cdot24=2,4\left(g\right)\) \(\Rightarrow m_{MgO}=4\left(g\right)\)
b) Ta có: \(\left\{{}\begin{matrix}n_{HCl\left(1\right)}=2n_{Mg}=0,2mol\\n_{HCl\left(2\right)}=2n_{MgO}=2\cdot\frac{4}{40}=0,2mol_{ }\end{matrix}\right.\)
\(\Rightarrow n_{HCl}=0,4mol\) \(\Rightarrow V_{ddHCl}=\frac{0,4}{1}=0,4\left(l\right)=400\left(ml\right)\)
c) Ta có: \(\left\{{}\begin{matrix}n_{MgCl_2\left(1\right)}=n_{Mg}=0,1mol\\n_{MgCl_2\left(2\right)}=n_{MgO}=0,1mol\end{matrix}\right.\)
\(\Rightarrow m_{muối}=m_{MgCl_2\left(1\right)}+m_{MgCl_2\left(2\right)}=0,1\cdot40+0,1\cdot40=8\left(g\right)\)
