Đáp án $+$ Giải thích các bước giải:

$n_{HCl}=0,4.2=0,8(mol)$

$n_{Al_2O_3}=\dfrac{10,2}{102}=0,1(mol)$

$2Na+2HCl→2NaCl+H_2$

$2Na+2H_2O→2NaOH+H_2$

$TH1:\ Na$ dư khi phản ứng $HCl$

$Al_2O_3+2NaOH→2NaAlO_2+H_2O$

Theo PT: $\Sigma_{Na}=n_{HCl}+2n_{Al_2O_3}=0,8+0,1.2=1(mol)$

$⇒m=1.23=23(g)$

$TH2:\ HCl$ dư 

$Al_2O_3+6HCl→2AlCl_3+3H_2O$

Theo PT: $n_{HCl\ (dư)}=6n_{Al_2O_3}=0,6(mol)$

$⇒n_{Na}=n_{HCl\ (pư)}=0,8-0,6=0,2(mol)$

$⇒m=0,2.23=4,6(g)$

Đổi `400(ml)=0,4(l)`

`=>n_(HCl)=0,4.2=0,8(mol)`

`n_(Al_2O_3)=(10,2)/102=0,1(mol)`

`TH_1:`

`PTHH:`

`(1)`

`2Na+2HCl->2NaCl+H_2`

`(2)`

`2Na+2H_2O->2NaOH+H_2`

`(3)`

`2NaOH+Al_2O_3->2NaAlO_2+H_2O`

Theo `PT(3),n_(NaOH)=2n_(Al_2O_3)=0,2(mol)`

Theo `PT(2),(3):`

`n_(Na)=n_(HCl)+n_(NaOH)=0,8+0,2`

`=>n_(Na)=1(mol)`

`=>m=m_(Na)=1.23=23(g)`

 `TH_2:`

`(3)`

`2Na+2HCl->2NaCl+H_2`

`(4)`

`Al_2O_3+6HCl->2AlCl_3+3H_2O`

`n_(HCl)(4)=6n_(Al_2O_3)=0,6(mol)`

`=>n_(HCl)(3)=0,8-0,6=0,2(mol)`

Theo `PT(3),n_(Na)=n_(HCl)=0,2(mol)`

`=>m=m_(Na)=0,2.23=4,6(g)`