Đáp án:

a. 

$\% {m_{Cu}} = 53,3\% $
$ \% {m_{Fe}} =  46,7\% $

b.

$m=  20,5$

$a=16$

c. 

$\begin{gathered} C{\% _{NaOH\,\,dư}} = 8,45\% ;\,\,C{\% _{N{a_2}S{O_3}}} = 15,14\% \hfill \\ {C_{M\,\,NaOH\,dư}} = 3,067M;\,\,{C_{M\,\,N{a_2}S{O_3}}} = 1,67M \hfill \\ \end{gathered} $

 Giải thích các bước giải:

\(\begin{array}{l}
Cu + 2{H_2}S{O_4} \to C{\rm{uS}}{O_4} + S{O_2} + 2{H_2}O\\
2Fe + 6{H_2}S{O_4} \to F{e_2}{{\rm{(S}}{O_4})_3} + 3S{O_2} + 6{H_2}O\\
{n_{S{O_2}}} = 0,25mol\\
a)
\end{array}\)

Gọi a và b lần lượt là số mol của Cu và Fe

\(\begin{array}{l}
\left\{ \begin{array}{l}
64a + 56b = 12\\
a + \dfrac{3}{2}b = 0,25
\end{array} \right. \to \left\{ \begin{array}{l}
a = 0,1\\
b = 0,1
\end{array} \right.\\
 \to {n_{Cu}} = 0,1mol \to {m_{Cu}} = 6,4g\\
 \to {n_{Fe}} = 0,1mol \to {m_{Fe}} = 5,6g\\
 \to \% {m_{Cu}} = \dfrac{{6,4}}{{12}} \times 100\%  = 53,3\% \\
 \to \% {m_{Fe}} = \dfrac{{5,6}}{{12}} \times 100\%  = 46,7\% \\
b)\\
C{\rm{uS}}{O_4} + 2KOH \to Cu{(OH)_2} + {K_2}S{O_4}\\
F{e_2}{{\rm{(S}}{O_4})_3} + 6KOH \to 2Fe{(OH)_3} + 3{K_2}S{O_4}\\
Cu{(OH)_2} \to CuO + {H_2}O\\
2Fe{(OH)_3} \to F{e_2}{O_3} + 3{H_2}O\\
{n_{Cu{{(OH)}_2}}} = {n_{{\rm{CuS}}{O_4}}} = {n_{Cu}} = 0,1mol
\end{array}\)

\(\begin{array}{l}
{n_{Fe{{(OH)}_3}}} = 2{n_{F{e_2}{{(S{O_4})}_3}}} = {n_{Fe}} = 0,1mol\\
 \to {m_{kếttủa}} = {m_{Cu{{(OH)}_2}}} + {m_{Fe\left( {OH} \right)}}_3 = 0,1 \times 98 + 0,1 \times 107 = 20,5\\
{n_{CuO}} = {n_{Cu{{(OH)}_2}}} = 0,1mol\\
{n_{F{e_2}{O_3}}} = \dfrac{1}{2}{n_{Fe{{(OH)}_3}}} = 0,05mol\\
 \to {m_{chấtrắn}} = {m_{CuO}} + {m_{F{e_2}{O_3}}} = 0,1 \times 80 + 0,05 \times 160 = 16g\\
c)\\
{m_{{\rm{dd}}NaOH}} = 150 \times 1,28 = 192g\\
 \to {m_{NaOH}} = \dfrac{{192 \times 20\% }}{{100\% }} = 38,4\\
 \to {n_{NaOH}} = 0,96mol\\
 \to \dfrac{{{n_{NaOH}}}}{{{n_{S{O_2}}}}} = \dfrac{{0,96}}{{0,25}} = 3,84
\end{array}\)

-> Tạo 1 muối: \(N{a_2}S{O_3}\)

\(\begin{array}{l}
S{O_2} + 2NaOH \to N{a_2}S{O_3} + {H_2}O\\
{n_{N{a_2}S{O_3}}} = {n_{S{O_2}}} = 0,25mol\\
 \to {m_{N{a_2}S{O_3}}} = 0,25 \times 126 = 31,5g\\
 \to {m_{{\rm{dd}}N{a_2}S{O_3}}} = {m_{S{O_2}}} + {m_{{\rm{dd}}NaOH}} = 0,25 \times 64 + 192 = 208g\\
 \to C{\% _{{\rm{dd}}N{a_2}S{O_3}}} = \dfrac{{31,5}}{{208}} \times 100\%  = 15,14\% 
\end{array}\)

$\to {C_{M\,\,N{a_2}S{O_3}}} = \dfrac{{0,25}}{{0,15}} = 1,67M$

$\begin{gathered} {n_{NaOH\,\,dư}} = 0,96 - 2.0,25 = 0,46\,\,mol\,\,\,\,\,\,\,\,\, \hfill \\ \to C{\% _{NaOH\,\,dư}} = \frac{{0,46.40}}{{208}}.100\% = 8,45\% \hfill \\ \to {C_{M\,\,NaOH}} = \frac{{0,46}}{{0,15}} = 3,067M\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \hfill \\ \end{gathered} $