Đáp án + Giải thích các bước giải:

$a\big)Fe_2O_3+3H_2\xrightarrow{t^o}2Fe+3H_2O$

$b\big)$

$n_{Fe_2O_3}=\frac{16}{160}=0,1(mol)$

Theo PT: $n_{H_2}=3n_{Fe_2O_3}=0,3(mol)$

$\to V_{H_2(đktc)}=0,3.22,4=6,72(l)$

$c\big)$

$n_{HCl}=\frac{125.14,6\%}{36,5}=0,5(mol)$

$n_{Fe}=2n_{Fe_2O_3}=0,2(mol)$

$Fe+2HCl\to FeCl_2+H_2$

Vì $n_{Fe}<\frac{n_{HCl}}{2}\to HCl$ dư

Theo PT: $n_{FeCl_2}=n_{H_2}=0,2(mol);n_{HCl(dư)}=0,5-0,2.2=0,1(mol)$

$C\%_{FeCl_2}=\frac{0,2.127}{0,2.56+125-0,2.2}.100\%\approx 18,7\%$

$C\%_{HCl(dư)}=\frac{0,1.36,5}{0,2.56+125-0,2.2}.100\%\approx 2,69\%$

Em tham khảo nha:

\(\begin{array}{l}
a)\\
F{e_2}{O_3} + 3{H_2} \xrightarrow{t^0} 2Fe + 3{H_2}O\\
b)\\
{n_{F{e_2}{O_3}}} = \dfrac{{16}}{{160}} = 0,1\,mol\\
{n_{{H_2}}} = 3{n_{F{e_2}{O_3}}} = 0,3\,mol\\
{V_{{H_2}}} = 0,3 \times 22,4 = 6,72l\\
c)\\
{n_{Fe}} = 2{n_{F{e_2}{O_3}}} = 0,2\,mol\\
{n_{HCl}} = \dfrac{{125 \times 14,6\% }}{{36,5}} = 0,5\,mol\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
{n_{Fe}} < \dfrac{{{n_{HCl}}}}{2} \Rightarrow \text{ HCl dư }  \\
{n_{FeC{l_2}}} = {n_{{H_2}}} = {n_{Fe}} = 0,2\,mol\\
{n_{HCl}} \text{ dư }= 0,5 - 0,2 \times 2 = 0,1\,mol\\
{m_{{\rm{dd}}spu}} = 125 + 0,2 \times 56 - 0,2 \times 2 = 135,8g\\
{C_\% }FeC{l_2} = \dfrac{{0,2 \times 127}}{{135,8}} \times 100\%  = 18,7\% \\
{C_\% }HCl  \text{  dư }= \dfrac{{0,1 \times 36,5}}{{135,8}} \times 100\%  = 2,69\% 
\end{array}\)