Đáp án+ Giải thích các bước giải:

`Pt : Fe + 2HCl->FeCl_2+H_2`

    `(0,2)->0,4->0,2->0,2\     (mol) `

`n_{Fe}={11,2}/56=0,2(mol)`

`a)`

`V_{H_2}=0,2.22,4=4,48(l)`

`m_{H_2}=0,2.2=0,4(g)`

`b)`

`m_{HCl}=0,4.36,5=14,6(g)`

`c)`

`C%_{dd\ HCl}={14,6.100%}/{200}=7,3%`

`d)`

`m_{FeCl_2}=0,2.127=25,4(g)`

\(\begin{array}{l}
a)\\
n_{Fe}=\frac{11,2}{56}=0,2(mol)\\
Fe+2HCl\to FeCl_2+H_2\\
Theo\,PT:\,n_{H_2}=n_{Fe}=0,2(mol)\\
\to m_{H_2}=0,2.2=0,4(g);V_{H_2}=0,2.22,4=4,48(l)\\
b)\\
Theo\,PT:\,n_{HCl}=2n_{H_2}=0,4(mol)\\
\to m_{HCl}=0,4.36,5=14,6(g)\\
c)\\
C\%_{HCl}=\frac{14,6}{200}.100\%=7,3\%\\
d)\\
Theo\,PT:\,n_{FeCl_2}=n_{Fe}=0,2(mol)\\
\to m_{FeCl_2}=0,2.127=25,4(g)
\end{array}\)