Đáp án:

`b.  m=11(g)`

Giải thích các bước giải:

`a.`

`Al+NaOH+H_2O->NaAlO_2+3/2H_2↑(1)`

`2Al+6HCl->2AlCl_3+3H_2↑(2)`

`Fe+2HCl->FeCl_2+H_2↑(3)`

`b.`

`n_(H_2(1))=(6,72)/(22,4)=0,3(mol)`

`=>n_(Al)=2/3n_(H_2(1))=0,2(mol)`

`=>n_(H_2(2))=3/2n_(Al)=0,3(mol)`

`=>n_(H_2(3))=(8,96)/(22,4)-0,3=0,1(mol)`

`=>n_(Fe)=n_(H_2(3))=0,1(mol)`

`m=m_(Al)+m_(Fe)=0,2xx27+0,1xx56=11(g)`

Đáp án:

$a)PTHH:$

$2Al+2NaOH+2H_2O\to 2NaAlO_2+3H_2\uparrow$

$2Al+6HCl\to 2AlCl_3+3H_2\uparrow$

$Fe+2HCl\to FeCl_2+H_2\uparrow$

$b)m=11$

Giải thích các bước giải:

$a)PTHH:$

$2Al+2NaOH+2H_2O\to 2NaAlO_2+3H_2\uparrow(1)$

$2Al+6HCl\to 2AlCl_3+3H_2\uparrow(2)$

$Fe+2HCl\to FeCl_2+H_2\uparrow(3)$

$n_{H_2(1)}=\dfrac{6,72}{22,4}=0,3(mol);n_{H_2(2+3)}=\dfrac{8,96}{22,4}=0,4(mol)$

Theo PT(1): $n_{Al}=\dfrac{2}{3}n_{H_2(1)}=0,2(mol)$

$\Rightarrow m_{Al}=0,2.27=5,4(g)$

Theo PT(2): $n_{H_2(2)}=\dfrac{3}{2}n_{Al}=0,3(mol)$

$\Rightarrow n_{H_2(1)}=0,4-0,3=0,1(mol)$

Theo PT(1): $n_{Fe}=n_{H_2(1)}=0,1(mol)$

$\Rightarrow m_{Fe}=0,1.56=5,6(g)$

$\Rightarrow m_{hh}=5,6+5,4=11(g)$

Vậy $m=11$