Đáp án:

\({m_{BaS{O_4}}} = 466{\text{ gam}}\)

\({\text{C}}{{\text{\% }}_{{H_2}S{O_4}}} = 78,4\% \)

\({{\text{m}}_{Fe{S_2}}} = 300{\text{ gam}}\)

Giải thích các bước giải:

Phản ứng xảy ra:

\(BaC{l_2} + {H_2}S{O_4}\xrightarrow{{}}BaS{O_4} + 2HCl\)

\({H_2}S{O_4} + 2KOH\xrightarrow{{}}{K_2}S{O_4} + 2{H_2}O\)

Ta có: 

\({n_{KOH}} = 4.1 = 4{\text{ mol}} \to {{\text{n}}_{{H_2}S{O_4}{\text{ dư}}}} = \frac{4}{2} = 2{\text{ mol}}\)

\({m_{BaC{l_2}}} = 4160.10\%  = 416{\text{ gam}} \to {{\text{n}}_{BaC{l_2}}} = {n_{BaS{O_4}}} = {n_{{H_2}S{O_4}}} = \frac{{416}}{{137 +  + 35,5.2}} = 2{\text{ mol}}\)

\( \to {m_{BaS{O_4}}} = 2.233 = 466{\text{ gam}}\)

\({n_{{H_2}S{O_4}}} = 2 + 2 = 4{\text{ mol}} \to {{\text{m}}_{{H_2}S{O_4}}} = 4.98 = 392{\text{ gam}} \to {\text{C}}{{\text{\% }}_{{H_2}S{O_4}}} = \frac{{392}}{{500}} = 78,4\% \)

\(4Fe{S_2} + 11{O_2}\xrightarrow{{{t^o}}}2F{e_2}{O_3} + 8S{O_2}\)

\({n_{{H_2}S{O_4}}} = 4 \to {n_{Fe{S_2}lt}} = \frac{1}{2}{n_{{H_2}S{O_4}}} = 2{\text{ mol}} \to {{\text{n}}_{Fe{S_2}}} = \frac{2}{{80\% }} = 2,5{\text{ mol}} \to {{\text{m}}_{Fe{S_2}}} = 2,5.120 = 300{\text{ gam}}\)