a) PTHH: \(MgO+2HCl\rightarrow MgCl_2+H_2O\)
b) Ta có: \(n_{MgO}=\frac{8}{40}=0,2\left(mol\right)\) \(\Rightarrow n_{HCl}=0,4mol\)
\(\Rightarrow m_{HCl}=0,4\cdot36,5=14,6\left(g\right)\)
\(\Rightarrow C\%_{HCl}=a\%=\frac{14,6}{200}\cdot100=7,3\%\)
c) Theo PTHH: \(n_{MgO}=n_{MgCl_2}=0,2mol\)
\(\Rightarrow m_{MgCl_2}=0,2\cdot95=19\left(g\right)\)
d) Ta có: \(m_{dd}=m_{MgO}+m_{ddHCl}=8+200=208\left(g\right)\)
\(\Rightarrow C\%_{MgCl_2}=\frac{19}{208}\cdot100\approx9,13\%\)