Đáp án:

\(\begin{array}{l}
b)\\
\% {m_{N{a_2}S{O_4}}} = 30,74\% \\
\% {m_{CuS{O_4}}} = 69,26\% \\
c)\\
{C_\% }NaCl = 2,03\% \\
{C_\% }CuC{l_2} = 4,685\% 
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
N{a_2}S{O_4} + BaC{l_2} \to 2NaCl + BaS{O_4}\\
CuS{O_4} + BaC{l_2} \to CuC{l_2} + BaS{O_4}\\
b)\\
hh:N{a_2}S{O_4}(a\,mol),CuS{O_4}(b\,mol)\\
{n_{BaS{O_4}}} = \dfrac{{34,95}}{{233}} = 0,15\,mol\\
\left\{ \begin{array}{l}
a + b = 0,15\\
142a + 160b = 23,1
\end{array} \right.\\
 \Rightarrow a = 0,05;b = 0,1\\
\% {m_{N{a_2}S{O_4}}} = \dfrac{{0,05 \times 142}}{{23,1}} \times 100\%  = 30,74\% \\
\% {m_{CuS{O_4}}} = 100 - 30,74 = 69,26\% \\
c)\\
{m_{{\rm{dd}}spu}} = 300 + 23,1 - 34,95 = 288,15g\\
{n_{NaCl}} = 2{n_{N{a_2}S{O_4}}} = 0,1\,mol\\
{n_{CuC{l_2}}} = {n_{CuS{O_4}}} = 0,1\,mol\\
{C_\% }NaCl = \dfrac{{0,1 \times 58,5}}{{288,15}} \times 100\%  = 2,03\% \\
{C_\% }CuC{l_2} = \dfrac{{0,1 \times 135}}{{288,15}} \times 100\%  = 4,685\% 
\end{array}\)