Đáp án:
\( {m_{BaS{O_4}}} = 466{\text{ gam}}\)
\( C{\% _{{H_2}S{O_4}}} = 78,4\% \)
\( {m_{Fe{S_2}}} = 300{\text{ gam}}\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(BaC{l_2} + {H_2}S{O_4}\xrightarrow{{}}BaS{O_4} + 2HCl\)
\(2KOH + {H_2}S{O_4}\xrightarrow{{}}{K_2}S{O_4} + 2{H_2}O\)
Ta có:
\({m_{BaC{l_2}}} = 4160.10\% = 416{\text{ gam}}\)
\( \to {n_{BaC{l_2}}} = \frac{{416}}{{137 + 35,5.2}} = 2{\text{ mol}}\)
\({n_{KOH}} = 4.1 = 4{\text{ mol}}\)
\({n_{BaS{O_4}}} = {n_{BaC{l_2}}} = 2{\text{ mol}}\)
\( \to {m_{BaS{O_4}}} = 0,2.233 = 466{\text{ gam}}\)
\({n_{{H_2}S{O_4}}} = {n_{BaC{l_2}}} + \frac{1}{2}{n_{KOH}} = 4{\text{ mol}}\)
\( \to {m_{{H_2}S{O_4}}} = 4.98 = 392{\text{ gam}}\)
\( \to C{\% _{{H_2}S{O_4}}} = \frac{{392}}{{500}} = 78,4\% \)
Sơ đồ:
\(Fe{S_2}\xrightarrow{{}}2S{O_2}\xrightarrow{{}}2S{O_3}\xrightarrow{{}}2{H_2}S{O_4}\)
Ta có:
\({n_{Fe{S_2}{\text{ lt}}}} = \frac{1}{2}{n_{{H_2}S{O_4}}} = 2{\text{ mol}}\)
\( \to {n_{Fe{S_2}}} = \frac{2}{{80\% }} = 2,5{\text{ mol}}\)
\( \to {m_{Fe{S_2}}} = 2,5.(56 + 32.2) = 300{\text{ gam}}\)
