Đáp án:
\({m_{rắn}} = 4,56{\text{ gam}}\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(2Na + 2{H_2}O\xrightarrow{{}}2NaOH + {H_2}\)
Ta có:
\({n_{Na}} = \frac{{9,2}}{{23}} = 0,4{\text{ mol = }}{{\text{n}}_{NaOH}}\)
\({m_{F{e_2}{{(S{O_4})}_3}}} = 200.4\% = 8{\text{ gam}} \to {{\text{n}}_{F{e_2}{{(S{O_4})}_3}}} = \frac{8}{{56.2 + 96.3}} = 0,02{\text{ mol; }}{{\text{m}}_{A{l_2}{{(S{O_4})}_3}}} = 6,84\% .200 = 13,68{\text{ gam}} \to {{\text{n}}_{A{l_2}{{(S{O_4})}_3}}} = \frac{{13,68}}{{27.2 + 96.3}} = 0,04{\text{ mol}}\)
Phản ứng xảy ra:
\(F{e_2}{(S{O_4})_3} + 6NaOH\xrightarrow{{}}2Fe{(OH)_3} + 3N{a_2}{(S{O_4})_3}\)
Ta có:
\({n_{Fe{{(OH)}_3}}} = 2{n_{F{e_2}{{(S{O_4})}_3}}} = 0,04{\text{ mol; }}{{\text{n}}_{NaOH{\text{ còn lại}}}} = 0,2 - 0,02.6 = 0,08{\text{ mol}}\)
\(A{l_2}{(S{O_4})_3} + 6NaOH\xrightarrow{{}}2Al{(OH)_3} + 3N{a_2}S{O_4}\)
\({n_{NaOH}} < 6{n_{A{l_2}{{(S{O_4})}_3}}} \to {n_{Al{{(OH)}_3}}} = \frac{1}{3}{n_{NaOH}} = \frac{{0,08}}{3}\)
Nung kết tủa:
\(2Fe{(OH)_3}\xrightarrow{{{t^o}}}F{e_2}{O_3} + 3{H_2}O\)
\(2Al{(OH)_3}\xrightarrow{{{t^o}}}A{l_2}{O_3} + 3{H_2}O\)
\( \to {n_{F{e_2}{O_3}}} = \frac{1}{2}{n_{Fe{{(OH)}_3}}} = 0,02{\text{ mol; }}{{\text{n}}_{A{l_2}{O_3}}} = \frac{1}{2}{n_{Al{{(OH)}_3}}} = \frac{{0,04}}{3} \to {m_{rắn}} = 0,02.160 + \frac{{0,04}}{3}.(27.2 + 16.3) = 4,56{\text{ gam}}\)
