1)

a)

Phản ứng xảy ra:

\(2Al + 3{H_2}S{O_4}\xrightarrow{{}}A{l_2}{(S{O_4})_3} + 3{H_2}\)

Ta có:

\({n_{{H_2}S{O_4}}} = 0,2.2 = 0,4{\text{ mol}}\)

\( \to {n_{Al}} = \frac{2}{3}{n_{{H_2}S{O_4}}} = \frac{2}{3}.0,4 = \frac{{0,8}}{3}{\text{ mol}}\)

\( \to {m_{Al}} = \frac{{0,8}}{3}.27 = 7,2{\text{ gam}}\)

b)

Phản ứng xảy ra:

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

\( \to {n_{NaOH}} = 2{n_{{H_2}S{O_4}}} = 0,2.2 = 0,4{\text{ mol}}\)

\( \to {m_{NaOH}} = 0,4.40 = 16{\text{ gam}}\)

\( \to {m_{dd\;{\text{NaOH}}}} = \frac{{{m_{NaOH}}}}{{C{\% _{NaOH}}}} = \frac{{16}}{{20\% }} = 80{\text{ gam}}\)

\( \to {V_{dd{\text{NaOH}}}} = \frac{{{m_{NaOH}}}}{d} = \frac{{80}}{{1,2}} = 66,67{\text{ ml}}\)

2)

Phản ứng xảy ra:

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

Rắn không tan là \(Cu\)

Ta có:

\({n_{{H_2}}} = \frac{{6,72}}{{22,4}} = 0,3{\text{ mol = }}{{\text{n}}_{Zn}} = {n_{ZnS{O_4}}} = {n_{{H_2}S{O_4}{\text{ phản ứng}}}}\)

\( \to {m_{Zn}} = 0,3.65 = 19,5{\text{ gam;}}{{\text{m}}_{Cu}} = 16{\text{ gam}}\)

\( \to \% {m_{Zn}} = \frac{{{m_{Zn}}}}{{{m_{Zn}} + {m_{Cu}}}} = \frac{{19,5}}{{19,5 + 16}} = 54,93\% \)

Ta có:

\({n_{{H_2}S{O_4}{\text{ dư}}}} = 0,3.20\%  = 0,06{\text{ mol}}\)

\({n_{{H_2}S{O_4}{\text{ tham gia}}}} = 0,3 + 0,06 = 0,36{\text{ mol}}\)

\( \to V = \frac{{0,36}}{2} = 0,18{\text{ lít}}\)

\( \to {C_{M{\text{ ZnS}}{{\text{O}}_4}}} = \frac{{0,3}}{{0,18}} = 1,67M;{C_{M{\text{ }}{{\text{H}}_2}S{O_4}{\text{ dư}}}} = \frac{{0,06}}{{0,18}} = 0,33M\)