Câu `1`

$\rm PTHH: MgO +2HCl \xrightarrow{} MgCl_2 +H_2O$ 

$\rm n_{MgO}=\dfrac{4}{40}=0,1(mol)$ 

theo pt 

$\rm n_{MgCl_2}=n_{Mg}=0,1(mol)$ 

`-.m_{MgCl_2}=0,1.95=95(g)` 

Câu `2` 

$\rm PTHH: SO_2 +2NaOH \xrightarrow{} Na_2SO_3 +H_2O$ 

$\rm n_{SO_2}=\dfrac{7,437}{24,79}=0,3(mol)$ 

theo pt

$\rm n_{Na_2SO_3}=n_{SO_2}=0,3(mol)$ 

$\rm → m_{Na_2SO_3}=0,3 .126=37,8(g)$ 

$1)\\ MgO+2HCl\to MgCl_2+H_2O\\ n_{MgCl_2}=n_{MgO}=\dfrac{4}{40}=0,1(mol)\\ m_{MgCl_2}=0,1.95=9,5(g)\\ 2)\\ 2NaOH+SO_2\to Na_2SO_3+H_2O\\ n_{Na_2SO_3}=n_{SO_2}=\dfrac{7,437}{24,79}=0,3(mol)\\ m_{Na_2SO_3}=0,3.126=37,8(g)$