Đáp án:

a, \({m_{BaS{O_4}}} = 466g\)

b, \(C{\% _{{H_2}S{O_4}}} = \dfrac{{196}}{{500}} \times 100\%  = 39,2\% \)

c, \({m_{Fe{S_2}}}thực= \dfrac{{120 \times 100}}{{80}} = 150g\)

Giải thích các bước giải:

\(\begin{array}{l}
BaC{l_2} + {H_2}S{O_4} \to BaS{O_4} + 2HCl\\
{n_{BaC{l_2}}} = \dfrac{{4160 \times 10}}{{100 \times 208}} = 2mol\\
{n_{HCl}} = 2{n_{BaC{l_2}}} = 4mol\\
HCl + KOH \to KCl + {H_2}O\\
{n_{KOH}} = 4mol = {n_{HCl}}
\end{array}\)

Vậy dung dịch B chỉ có axit HCl

\(\begin{array}{l}
{n_{BaS{O_4}}} = {n_{BaC{l_2}}} = 2mol\\
 \to {m_{BaS{O_4}}} = 466g\\
{n_{{H_2}S{O_4}}} = {n_{BaC{l_2}}} = 2mol\\
 \to {m_{{H_2}S{O_4}}} = 196g\\
 \to C{\% _{{H_2}S{O_4}}} = \dfrac{{196}}{{500}} \times 100\%  = 39,2\% \\
Fe{S_2} \to S{O_2} \to S{O_3} \to {H_2}S{O_4}\\
Fe{S_2} \to 2{H_2}S{O_4}\\
{n_{Fe{S_2}}} = \dfrac{1}{2}{n_{{H_2}S{O_4}}} = 1mol\\
 \to {m_{Fe{S_2}}} = 120g\\
 \to {m_{Fe{S_2}}}thực= \dfrac{{120 \times 100}}{{80}} = 150g
\end{array}\)