Đáp án:

\( {V_{{H_2}}} = 4,48{\text{ lít}}\)

\( V = {V_{HCl}}= 0,2{\text{ lít}}\)

\({m_{dd\;{{\text{H}}_2}S{O_4}}}  = 80{\text{ gam}}\)

Giải thích các bước giải:

 Phản ứng xảy ra:

\(Zn + 2HCl\xrightarrow{{}}ZnC{l_2} + {H_2}\)

Ta có:

\({n_{Zn}} = \frac{{13}}{{65}} = 0,2{\text{ mol = }}{{\text{n}}_{{H_2}}}\)

\( \to {V_{{H_2}}} = 0,2.22,4 = 4,48{\text{ lít}}\)

\( \to {n_{HCl}} = 2{n_{{H_2}}} = 0,2.2 = 0,4{\text{ mol}}\)

\( \to V = {V_{HCl}} = \frac{{0,4}}{2} = 0,2{\text{ lít}}\)

Thay \(HCl\) bằng \(H_2SO_4\)

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

Ta có:

\({n_{{H_2}S{O_4}}} = {n_{Zn}} = 0,2{\text{ mol}}\)

\( \to {m_{{H_2}S{O_4}}} = 0,2.98 = 19,6{\text{ gam}}\)

\({m_{dd\;{{\text{H}}_2}S{O_4}}} = \frac{{19,6}}{{24,5\% }} = 80{\text{ gam}}\)