Đáp án:

 a) C% BaCl2 dư=5,39%; C% FeCl3=8,41%

b) m=8 gam

Giải thích các bước giải:

Ta có: \({m_{BaC{l_2}}} = 208.20\%  = 41,6{\text{ gam}} \to {{\text{n}}_{BaC{l_2}}} = \frac{{41,6}}{{137 + 35,5.2}} = 0,2{\text{ mol; }}{{\text{n}}_{F{e_2}{{(S{O_4})}_3}}} = \frac{{20}}{{56.2 + 96.3}} = 0,05{\text{ mol}}\)

\(3BaC{l_2} + F{e_2}{(S{O_4})_3}\xrightarrow{{}}3BaS{O_4} + 2FeC{l_3}\)

 Vì \({n_{BaC{l_2}}} > 3{n_{F{e_2}{{(S{O_4})}_3}}}\) nên BaCl2 dư.

\( \to {n_{BaS{O_4}}} = 0,15{\text{ mol; }}{{\text{n}}_{BaC{l_2}du}} = 0,2 - 0,15 = 0,05{\text{ mol; }}{{\text{n}}_{FeC{l_3}}} = 2{n_{F{e_2}{{(S{O_4})}_3}}} = 0,1{\text{ mol}}\)

\( \to BTKL:{\text{ }}{{\text{m}}_{dd{\text{ sau phản ứng}}}} = 208 + 20 - {m_{BaS{O_4}}} = 208 + 20 - 0.15.233 = 193,05{\text{ gam}}\)

\({m_{BaC{l_2}{\text{ dư}}}} = 0,05.(137 + 35,5.2) = 10,4{\text{ gam; }}{{\text{m}}_{FeC{l_3}}} = 0,1.(56 + 35,5.3) = 16,25{\text{ gam}}\)

\[ \to C{\% _{BaC{l_2}{\text{ dư}}}} = \frac{{10,4}}{{193,05}} = 5,39\% ;{\text{ C}}{{\text{\% }}_{FeC{l_3}}} = \frac{{16,25}}{{193,05}} = 8,41\% \)

Cho NaOH vào dung dịch B.

\(FeC{l_3} + 3NaOH\xrightarrow{{}}Fe{(OH)_3} + 3NaCl\)

\( \to {n_{Fe{{(OH)}_3}}} = {n_{FeC{l_3}}} = 0,1{\text{ mol}}\)

\(2Fe{(OH)_3}\xrightarrow{{}}F{e_2}{O_3} + 3{H_2}O\)

\( \to {n_{F{e_2}{O_3}}} = \frac{1}{2}{n_{Fe{{(OH)}_3}}} = 0,05{\text{ mol}} \to {{\text{m}}_{F{e_2}{O_3}}} = 0,05.(56.2 + 16.3) = 8{\text{ gam}}\)