Em tham khảo nha :

\(\begin{array}{l}
17)\\
a)\\
FeC{l_2} + 2NaOH \to Fe{(OH)_2} + 2NaCl\\
4Fe{(OH)_2} + {O_2} \to 2F{e_2}{O_3} + 4{H_2}O\\
b)\\
{n_{FeC{l_2}}} = 0,15 \times 0,2 = 0,03mol\\
{n_{Fe{{(OH)}_2}}} = {n_{FeC{l_2}}} = 0,03mol\\
{n_{F{e_2}{O_3}}} = \dfrac{{{n_{Fe{{(OH)}_2}}}}}{2} = 0,015mol\\
{m_{F{e_2}{O_3}}} = 0,015 \times 160 = 2,4g\\
c)\\
{n_{NaCl}} = 2{n_{FeC{l_2}}} = 0,06mol\\
{C_{{M_{NaCl}}}} = \dfrac{{0,06}}{{0,2 + 0,3}} = 0,12M\\
18)\\
a)\\
Zn + CuS{O_4} \to ZnS{O_4} + Cu\\
\text{Phản ứng này là phản ứng thế}\\
b)\\
{m_{CuS{O_4}}} = \dfrac{{32 \times 10}}{{100}} = 3,2g\\
{n_{CuS{O_4}}} = \dfrac{{3,2}}{{160}} = 0,02mol\\
{n_{Zn}} = {n_{CuS{O_4}}} = 0,02mol\\
{m_{Zn}} = 0,02 \times 65 = 1,3g\\
c)\\
{n_{Cu}} = {n_{CuS{O_4}}} = 0,02mol\\
{m_{Cu}} = 0,02 \times 64 = 1,28g\\
{n_{ZnS{O_4}}} = {n_{Zn}} = 0,02mol\\
{m_{ZnSO4}} = 0,02 \times 161 = 3,22g\\
{m_{{\rm{dd}}spu}} = 1,3 + 30 - 1,28 = 30,02g\\
C{\% _{ZnS{O_4}}} = \dfrac{{3,22}}{{30,02}} \times 100\%  = 10,7\% 
\end{array}\)