Đáp án:
\(a\%=18,375\%\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(Fe + {H_2}S{O_4}\xrightarrow{{}}FeS{O_4} + {H_2}\)
\(A{l_2}{O_3} + 3{H_2}S{O_4}\xrightarrow{{}}A{l_2}{(S{O_4})_3} + 3{H_2}O\)
Ta có:
\({n_{{H_2}}} = \frac{{3,36}}{{22,4}} = 0,15{\text{ mol}}\)
\( \to {n_{Fe}} = {n_{{H_2}}} = 0,15{\text{ mol}}\)
\( \to {m_{Fe}} = 0,15.56 = 8,4{\text{ gam}}\)
\( \to {n_{A{l_2}{O_3}}} = 28,8 - 8,4 = 20,4{\text{ gam}}\)
\({n_{A{l_2}{O_3}}} = \frac{{20,4}}{{27.2 + 16.3}} = 0,2{\text{ mol}}\)
\(\to {n_{{H_2}S{O_4}}} = {n_{Fe}} + 3{n_{A{l_2}{O_3}}} = 0,15 + 0,2.3 = 0,75{\text{ mol}}\)
\( \to {m_{{H_2}S{O_4}}} = 0,75.98 = 73,5{\text{ gam}}\)
\( \to a\% = \frac{{73,5}}{{400}} = 18,375\% \)
BTKL:
\({m_{dd{\text{ sau phản ứng}}}} = {m_{hh}} + {m_{dd{\text{ }}{{\text{H}}_2}S{O_4}}} - {m_{{H_2}}}\)
\( = 28,8 + 400 - 0,15.2 = 428,5{\text{ gam}}\)
\({n_{FeS{O_4}}} = {n_{Fe}} = 0,15{\text{ mol}} \to {{\text{m}}_{FeS{O_4}}} = 0,15.(56 + 96) = 22,8{\text{ gam}}\)
\({n_{A{l_2}{{(S{O_4})}_3}}} = {n_{A{l_2}{O_3}}} = 0,2{\text{ mol}} \to {{\text{m}}_{A{l_2}{{(S{O_4})}_3}}} = 0,2.(27.2 + 96.3) = 68,4{\text{ gam}}\)
\( \to C{\% _{FeS{O_4}}} = \frac{{22,8}}{{428,5}} = 5,32\% \)
\(C{\% _{A{l_2}{{(S{O_4})}_3}}} = \frac{{68,4}}{{428,5}} = 15,96\% \)
