Đáp án:

 a) C% H2SO4=16%

b) C% ZnSO4=18,17%; C% H2SO4 dư=3,69%

Giải thích các bước giải:

({n_{ZnO}} = \frac{{8,1}}{{65 + 16}} = 0,1{\text{ mol; }}{{\text{n}}_{Zn}} = \frac{{13}}{{65}} = 0,2{\text{ mol; }}{{\text{n}}_{NaOH}} = 0,05.4 = 0,2{\text{ mol}}\)

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

\(ZnO + {H_2}S{O_4}\xrightarrow{{}}ZnS{O_4} + {H_2}O\)

\(2NaOH + {H_2}S{O_4}\xrightarrow{{}}N{a_2}S{O_4} + 2{H_2}O\)

\( \to {n_{{H_2}S{O_4}}} = {n_{Zn}} + {n_{ZnO}} + \frac{1}{2}{n_{NaOH}} = 0,2 + 0,1 + \frac{{0,2}}{2} = 0,4{\text{ mol}} \to {{\text{m}}_{{H_2}S{O_4}}} = 0,4.98 = 39,2{\text{ gam}} \to {\text{C}}{{\text{\% }}_{{H_2}S{O_4}}} = \frac{{39,2}}{{245}} = 16\% \)

\({n_{{H_2}}} = {n_{Zn}} = 0,2{\text{ mol}} \to {\text{BTKL: }}{{\text{m}}_Y} = {m_{Zn}} + {m_{ZnO}} + {m_{dd\;{{\text{H}}_2}S{O_4}}} - {m_{{H_2}}} = 8,1 + 13 + 245 - 0,2.2 = 265,7{\text{ gam}}\)

Dung dịch Y chứa ZnSO4 0,3 mol và H2SO4 dư 0,1 mol.

\( \to {m_{ZnS{O_4}}} = 0,3.(65 + 96) = 48,3{\text{ gam}} \to C{\% _{ZnS{O_4}}} = \frac{{48,3}}{{265,7}} = 18,17\% \)

\({m_{{H_2}S{O_4}{\text{ du}}}} = 0,1.98 = 9,8{\text{ gam}} \to C{\% _{{H_2}S{O_4}{\text{ du}}}} = \frac{{9,8}}{{265,7}} = 3,69\% \)