1)

Phản ứng xảy ra:

\(Zn + {H_2}S{O_4}\xrightarrow{{}}ZnS{O_4} + {H_2}\)

 Ta có:

\({n_{Zn}} = \frac{{9,75}}{{65}} = 0,15{\text{ mol = }}{{\text{n}}_{{H_2}S{O_4}}} = {n_{ZnS{O_4}}} = {n_{{H_2}}}\)

\( \to {V_{{H_2}}} = 0,15.22,4 = 3,36{\text{ lít}}\)

\( \to {m_{ZnS{O_4}}} = 0,15.(65 + 96) = 24,15{\text{ gam}}\)

Ta có:

\({n_{{H_2}S{O_4}{\text{ dư}}}} = 0,15.15\%  = 0,0225{\text{ mol}}\)

\( \to {n_{{H_2}S{O_4}{\text{ tham gia}}}} = 0,15 + 0,0225 = 0,1725{\text{ mol}}\)

2)

Phản ứng xảy ra:

\(Ba{(OH)_2} + {H_2}S{O_4}\xrightarrow{{}}BaS{O_4} + 2{H_2}O\)

Ta có:

\({n_{{H_2}S{O_4}}} = 0,2.0,5 = 0,1{\text{ mol = }}{{\text{n}}_{Ba{{(OH)}_2}}} = {n_{BaS{O_4}}}\)

\( \to V = {V_{dd{\text{ Ba(OH}}{{\text{)}}_2}}} = \frac{{0,1}}{{0,5}} = 0,2{\text{ lít = 200 ml}}\)

\({m_{BaS{O_4}}} = 0,1.(137 + 96) = 23,3{\text{ gam}}\)