Đáp án:

\(\begin{array}{l}
a)\\
{m_{KX}} = 46,15g\\
b)\\
Clo(Cl),Brom(Br)\\
\% {m_{BaC{l_2}}} = 51,23\% \\
\% {m_{BaB{r_2}}} = 48,77\% 
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
Ba{X_2} + {K_2}S{O_4} \to BaS{O_4} + 2KX\\
{n_{BaS{O_4}}} = \dfrac{{58,25}}{{233}} = 0,25\,mol\\
{n_{{K_2}S{O_4}}} = {n_{BaS{O_4}}} = 0,25\,mol\\
{m_{KX}} = 60,9 + 0,25 \times 174 - 58,25 = 46,15g\\
b)\\
{n_{Ba{X_2}}} = {n_{BaS{O_4}}} = 0,25\,mol\\
{M_{Ba{X_2}}} = \dfrac{{60,9}}{{0,25}} = 243,6(g/mol)\\
 \Rightarrow {M_X} = 53,3(g/mol)\\
 \Rightarrow Clo(Cl),Brom(Br)\\
hh:BaC{l_2}(a\,mol),BaB{r_2}(b\,mol)\\
a + b = 0,25\\
208a + 297b = 60,9\\
 \Rightarrow a = 0,15;b = 0,1\\
\% {m_{BaC{l_2}}} = \dfrac{{0,15 \times 208}}{{60,9}} \times 100\%  = 51,23\% \\
\% {m_{BaB{r_2}}} = 100 - 51,23 = 48,77\% 
\end{array}\)