Đáp án:

 a) H2SO4 dư 19,6 gam

b) m dung dịch BaCl2=1248 gam

Giải thích các bước giải:

 Phản ứng: \(A{l_2}{O_3} + 3{H_2}S{O_4}\xrightarrow{{}}A{l_2}{(S{O_4})_3} + 3{H_2}O\)

Ta có:  \({n_{A{l_2}{O_3}}} = \frac{{10,2}}{{27.2 + 16.3}} = 0,1{\text{ mol; }}{{\text{n}}_{{H_2}S{O_4}}} = \frac{{49}}{{98}} = 0,5{\text{ mol}}\)

Vì \({n_{{H_2}S{O_4}}} > 3{n_{A{l_2}{O_3}}}\) nên H2SO4 dư.

\(\to {n_{{H_2}S{O_4}{\text{ dư}}}} = 0,5 - 0,1.3 = 0,2{\text{ mol}} \to {{\text{m}}_{{H_2}S{O_4}{\text{ dư}}}} = 0,2.98 = 19,6{\text{ gam}}\)

Ta có:  \({n_{A{l_2}{{(S{O_4})}_3}}} = {n_{A{l_2}{O_3}}} = 0,1{\text{ mol}}\)

\(A{l_2}{(S{O_4})_3} + 3BaC{l_2}\xrightarrow{{}}3BaS{O_4} + 2AlC{l_3}\)

\(\to {n_{BaC{l_2}}} = 3{n_{A{l_2}{{(S{O_4})}_3}}} = 0,1.3 = 0,3{\text{ mol}} \to {{\text{m}}_{BaC{l_2}}} = 0,3.(137 + 35,5.2) = 62,4{\text{ gam}} \to {{\text{m}}_{dd\;{\text{BaC}}{{\text{l}}_2}}} = \frac{{62,4}}{{5\% }} = 1248{\text{ gam}}\)