\(\begin{array}{l}
a)\\
n_{H_2}=\frac{6,72}{22,4}=0,3(mol)\\
2K+2H_2O\to 2KOH+H_2\\
n_K=2n_{H_2}=0,6(mol)\\
m=m_K+m_{FeO}=0,6.39+14,4=37,8(g)\\
b)\\
n_{KOH}=n_K=0,6(mol)\\
C_{M\,KOH}=\frac{0,6}{2}=0,3M\\
c)\\
n_{HCl}=\frac{182,5.10\%}{36,5}=0,5(mol)\\
n_{FeO}=\frac{14,4}{72}=0,2(mol)\\
FeO+2HCl\to FeCl_2+H_2O\\
Vi\,n_{HCl}:2>n_{FeO}\to HCl\,du\,0,5-0,2.2=0,1(mol)\\
n_{FeCl_2}=n_{FeO}=0,2(mol)\\
C\%_{FeCl_2}=\frac{0,2.127}{14,4+182,5}.100\%≈13\%\\
C\%_{HCl(du)}=\frac{0,1.36,5}{14,4+182,5}.100\%≈1,85\%
\end{array}\)

Em tham khảo nha :

\(\begin{array}{l}
a)\\
2K + 2{H_2}O \to 2KOH + {H_2}\\
{n_{{H_2}}} = \dfrac{{6,72}}{{22,4}} = 0,3mol\\
{n_K} = 2{n_{{H_2}}} = 0,6mol\\
{m_K} = 0,6 \times 39 = 23,4g\\
m = 23,4 + 14,4 = 47,8g\\
b)\\
{n_{KOH}} = {n_K} = 0,6mol\\
{C_{{M_{KOH}}}} = \dfrac{{0,6}}{2} = 0,3M\\
c)\\
FeO + 2HCl \to FeC{l_2} + {H_2}O\\
{n_{FeO}} = \dfrac{{14,4}}{{72}} = 0,2mol\\
{n_{HCl}} = \dfrac{{182,5 \times 10\% }}{{36,5}} = 0,5mol\\
\dfrac{{0,2}}{1} < \dfrac{{0,5}}{2} \Rightarrow HCl\text{ dư}\\
{n_{HC{l_d}}} = 0,5 - 0,2 \times 2 = 0,1mol\\
{n_{FeC{l_2}}} = {n_{FeO}} = 0,2mol\\
C{\% _{HCl}} = \dfrac{{0,1 \times 36,5}}{{14,4 + 182,5}} \times 100\%  = 1,85\% \\
C{\% _{FeC{l_2}}} = \dfrac{{0,2 \times 127}}{{14,4 + 182,5}} \times 100\%  = 12,9\% 
\end{array}\)