1)

Phản ứng xảy ra:

\(Fe + 2HCl\xrightarrow{{}}FeC{l_2} + {H_2}\)

Ta có:

\({n_{Fe}} = \frac{{11,2}}{{56}} = 0,2{\text{ mol}}\)

Theo phản ứng:

\({n_{Fe}} = {n_{{H_2}}} = 0,2{\text{ mol;}}{{\text{n}}_{HCl}} = 2{n_{Fe}} = 0,2.2 = 0,4{\text{ mol}}\)

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

\( \to {V_{HCl}} = \frac{{0,4}}{1} = 0,4{\text{ lít}}\)

2)

Phản ứng xảy ra:

\(N{a_2}C{O_3} + 2HCl\xrightarrow{{}}2NaCl + C{O_2} + {H_2}O\)

Ta có:

\({n_{N{a_2}C{O_3}}} = \frac{{42,4}}{{23.2 + 60}} = 0,4{\text{ mol}}\)

\( \to {n_{C{O_2}}} = {n_{N{a_2}C{O_3}}} = 0,4{\text{ mol}}\)

\({n_{HCl}} = 2{n_{N{a_2}C{O_3}}} = 0,4.2 = 0,8{\text{ mol}}\)

\( \to {V_{C{O_2}}} = 0,4.22,4 = 8,96{\text{ lít}}\)

\( \to {C_{M{\text{ HCl}}}} = \frac{{0,8}}{{0,1}} = 8M\)

3)

Phản ứng xảy ra:

\({K_2}C{O_3} + 2HCl\xrightarrow{{}}2KCl + C{O_2} + {H_2}O\)

Ta có:

\({n_{{K_2}C{O_3}}} = \frac{{42,4}}{{39.2 + 60}} = 0,3072 = {n_{C{O_2}}}\)

\( \to {V_{C{O_2}}} = 0,3072.22,4 = 6,88128{\text{ lít}}\)

\({n_{HCl}} = 2{n_{{K_2}C{O_3}}} = 0,3072.2 = 0,6144{\text{ mol}}\)

\( \to {C_{M{\text{ HCl}}}} = \frac{{0,6144}}{{0,1}} = 6,144M\)