$1)$

$a)PTHH:$

$2Na+2H_2O \to 2NaOH+H_2$

$b)$

`n_(Na)=(4,6)/23=0,2(mol)`

Theo `PT,n_(NaOH)=n_(Na)=0,2(mol)`

`=>m_(NaOH)=0,2.40=8(g)`

$c)$

Theo `PT,n_(H_2)=1/2n_(Na)=0,1(mol)`

`=>V_(H_2)(đktc)=0,1.22,4=2,24(l)`

$d)$

`n_(CuO)=12/80=0,15(mol)`

$PT:H_2+CuO\xrightarrow{t^0}Cu+H_2O$

Xét tỉ lệ :

`(n_(CuO))/1=0,15>0,1=(n_(H_2))/1`

`=>CuO` dư,$H_2$ hết

`=>Mol` tính theo $H_2$

Theo `PT,n_(CuO)(pứ)=n_(H_2)=0,1(mol)`

 `=>m_(CuO)(pứ)=0,1.80=8(g)`

`=>m_(CuO)(dư)=12-8=4(g)`

$2)$

$a)PTHH:$

$Mg+2HCl \to MgCl_2+H_2$

$b)$

`n_(Mg)=(2,4)/24=0,1(mol)`

Theo `PT,n_(H_2)=n_(Mg)=0,1(mol)`

`=>V_(H_2)(đktc)=0,1.22,4=2,24(l)`

$c)$

Theo `PT,n_(HCl)=2n_(Mg)=0,2(mol)`

`=>m_(ctHCl)=0,2.36,5=7,3(g)`

`=>m_(ddHCl)=(7,3.100)/(14,6)`

`=>m_(ddHCl)=50(g)`