1)

Phản ứng xảy ra:

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

Ta có:

\({n_{Zn}} = \frac{{6,5}}{{65}} = 0,1{\text{ mol}}\)

Theo phản ứng:

\({n_{HCl}} = 2{n_{Zn}} = 0,1.2 = 0,2{\text{ mol }}\)

\({n_{{H_2}}} = {n_{Zn}} = 0,1{\text{ mol}} \to {{\text{V}}_{{H_2}}} = 0,1.22,4 = 2,24{\text{ lít}}\)

Trung hòa dung dịch \(HCl\) trên bằng \(NaOH\).

\(NaOH + HCl\xrightarrow{{}}NaCl + {H_2}O\)

Ta có:

\({n_{HCl}} = {n_{NaOH}} = 0,2{\text{ mol}} \to {{\text{V}}_{NaOH}} = \frac{{0,2}}{2} = 0,1{\text{ lít}}\)

2)

Phản ứng xảy ra:

\(Cu + 2AgN{O_3}\xrightarrow{{}}Cu{(N{O_3})_2} + 2Ag\)

\(Ag\) không phản ứng nên rắn còn lại là \(Ag\).

\( \to {m_{Ag}} = 10,8{\text{ gam}} \to {{\text{m}}_{Cu}} = 12,8 - 10,8 = 2{\text{ gam}}\)

\( \to \% {m_{Ag}} = \frac{{10,8}}{{12,8}} = 84,375\%  \to \% {m_{Cu}} = 15,625\% \)

Ta có:

\({n_{Cu}} = \frac{2}{{64}} = \frac{1}{{32}}{\text{ mol}} \to {{\text{n}}_{AgN{O_3}}} = 2{n_{Cu}} = \frac{1}{{16}}{\text{ mol}}\)

\( \to {V_{dd{\text{AgN}}{{\text{O}}_3}}} = \frac{{\frac{1}{{16}}}}{2} = \frac{1}{{32}}{\text{ lít}}\)