Em tham khảo nha:

 \(\begin{array}{l}
a)\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
b)\\
{n_{Fe}} = \dfrac{{2,8}}{{56}} = 0,05\,mol\\
{n_{HCl}} = \dfrac{{14,6}}{{36,5}} = 0,4\,mol\\
\dfrac{{0,05}}{1} < \dfrac{{0,4}}{2} \Rightarrow HCl \text{ dư } \\
{n_{HCl}}  \text{ dư }= 0,4 - 0,05 \times 2 = 0,3\,mol\\
{m_{HCl}}  \text{ dư }= 0,3 \times 36,5 = 10,95g\\
c)\\
{n_{{H_2}}} = {n_{Fe}} = 0,05\,mol\\
{V_{{H_2}}} = 0,05 \times 22,4 = 1,12l\\
d)\\
{n_{Fe}} = \dfrac{{{n_{HCl}  \text{ dư }}}}{2} = \dfrac{{0,3}}{2} = 0,15\,mol\\
{m_{Fe}} = 0,15 \times 56 = 8,4g
\end{array}\)

$n_{Fe}=\frac{m}{M}=0,05(mol)$

$n_{HCl}=\frac{m}{M}=0,4(mol)$

=>$Fe+2HCl\rightarrow FeCl_2+ H_2$

=>Fe hết, HCl dư

$n_{HCl_{pư}}=0,1(mol)$

=>$n_{HCl} dư=0,3(mol)$

Vậy $m_{HCl} dư=10,95(g)$

c, $n_{H_2}=n_{Fe}=0,05(mol)$

=> $V_{H_2}=n.22,4=1,12(l)$

d, $n_{Fe}=\frac{1}{2}. n_{HCl}=0,2(mol)$

=>$m_{Fe}=n.M=11,2(g)$

=>$m_{Fe} thêm=8,4(g)$