`4,`

`n_{H_2}=(1,12)/(22,4)=0,05(mol)`

PTHH:

`Fe+2HCl->FeCl_2+H_2`

       `0,1`              `←0,05(mol)`

`a,`

`V_{dd HCl}=(0,1)/1=0,1(lít)` hay `100ml`

`b,`

PTHH:

`Fe+H_2SO_4->FeSO_4+H_2`

      `0,05`                       `←0,05(mol)`

`=>V_{dd H_2SO_4}=(0,05)/(0,5)=0,1(lít)` hay `100ml`

`5,`

`n_{Al}=(5,4)/(27)=0,2(mol)`

PTHH:

`2Al+3H_2SO_4->Al_2(SO_4)_3+3H_2`

`0,2->0,3(mol)`

`a,`

`m_{H_2SO_4}=0,3.98=29,4(g)`

`=>m_{dd H_2SO_4(pứ)}=(29,4)/(10\%)=294(g)`

`b,`

PTHH:

`2Al+6HCl->2AlCl_3+3H_2`

`0,2->0,6(mol)`

`=>V_{dd HCl}=(0,6)/(2)=0,3(lít)` 

Giải thích các bước giải:

\(\begin{array}{l} 4.\\a)\\n_{H_2}=\dfrac{1,12}{22,4}=0,05(mol)\\Fe+2HCl\to FeCl_2+H_2\\n_{HCl}=2n_{H_2}=0,1(mol)\\V_{HCl}=\dfrac{0,1}{1}=0,1(l)\\b)\\Fe+H_2SO_4\to FeSO_4+H_2\\n_{H_2SO_4}=n_{H_2}=0,05(mol)\\V_{H_2SO_4}=\dfrac{0,05}{0,5}=0,1(mol)\\5.\\n_{Al}=\dfrac{5,4}{27}=0,2(mol)\\2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2\\n_{H_2SO_4}=\dfrac{3}{2}n_{Al}=0,3(mol)\\m_{dd\ H_2SO_4}=\dfrac{0,3×98}{10\%}=294(g)\\b)\\2Al+6HCl\to 2AlCl_3+3H_2\\n_{HCl}=3n_{Al}=0,6(mol)\\V_{HCl}=\dfrac{0,6}{2}=0,3(l)\end{array}\)