1)

Phản ứng xảy ra:

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

Ta có: 

\({n_{Zn}} = {n_{{H_2}}} = \frac{{11,2}}{{22,4}} = 0,5{\text{ mol}} \to {{\text{m}}_{Zn}} = m = 0,5.65 = 32,5{\text{ gam}}\)

Ta có: 

\({n_{ZnC{l_2}}} = {n_{Zn}} = 0,5{\text{ mol}} \to {{\text{m}}_{ZnC{l_2}}} = 0,5.(65 + 35,5.2) = 68{\text{ gam}}\)

\({n_{HCl}} = 2{n_{{H_2}}} = 1{\text{ mol}} \to {{\text{m}}_{HCl}} = 1.36,5 = 36,5{\text{ gam}} \to {m_{dd{\text{ HCl}}}} = \frac{{36,5}}{{25\% }} = 146{\text{ gam}}\)

2)

Phản ứng xảy ra:

\(Mg + {H_2}S{O_4}\xrightarrow{{}}MgS{O_4} + {H_2}\)

Ta có: 

\({n_{Mg}} = {n_{{H_2}S{O_4}}} = {n_{MgS{O_4}}} = {n_{{H_2}}} = \frac{{6,72}}{{22.4}} = 0,3{\text{ mol}}\)

\( \to m = {m_{Mg}} = 0,3.24 = 7,2{\text{ gam}}\)

\({m_{MgS{O_4}}} = 0,3.(24 + 96) = 360{\text{ gam}}\)

\({m_{{H_2}S{O_4}}} = 0,3.98 = 29,4{\text{ gam}} \to {m_{dd{\text{ }}{{\text{H}}_2}S{O_4}}} = \frac{{29,4}}{{25\% }} = 117,6{\text{ gam}}\)

3)

Phản ứng xảy ra:

\(2Al + 6HCl\xrightarrow{{}}2AlC{l_3} + 3{H_2}\)

Ta có: \({n_{HCl}} = 0,3.0,2 = 0,06{\text{ mol}}\)

\( \to {n_{Al}} = {n_{AlC{l_3}}} = \frac{1}{3}{n_{HCl}} = 0,02{\text{ mol; }}{{\text{n}}_{{H_2}}} = \frac{1}{2}{n_{HCl}} = 0,03{\text{ mol}}\)

\(m = {m_{Al}} = 0,02.27 = 0,54{\text{ gam}}\)

\(V = {V_{{H_2}}} = 0,03.22,4 = 0,672{\text{ lít}}\)

\({m_{AlC{l_3}}} = 0,02.(27 + 35,5.3) = 2,67{\text{ gam}}\)

\({V_{dd}} = {V_{dd{\text{ HCl}}}} = 300ml = 0,3{\text{ lít}}\)

\( \to {C_{M{\text{ AlC}}{{\text{l}}_3}}} = \frac{{0,02}}{{0,3}} = 0,0667M\)