2)

Phản ứng xảy ra:

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

Ta có:

\({n_{Fe}} = \frac{{{m_{Fe}}}}{{{M_{Fe}}}} = \frac{7}{{56}} = 0,125{\text{ mol}}\)

Theo tỉ lệ phản ứng:

\({n_{{H_2}}} = {n_{Fe}} = 0,125{\text{ mol}} \to {V_{{H_2}}} = 0,125.22,4 = 2,8{\text{ lít}}\)

\({n_{HCl}} = 2{n_{Fe}} = 0,125.2 = 0,25{\text{ mol}}\)

\( \to {m_{HCl}} = 0,25.36,5 = 9,125{\text{ gam}}\)

\( \to {m_{dd\;{\text{HCl}}}} = \frac{{{m_{HCl}}}}{{C{\% _{HCl}}}} = \frac{{9,125}}{{9,125\% }} = 100{\text{ gam}}\)

3)

Phản ứng xảy ra:

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

Ta có:

\({m_{{H_2}S{O_4}}} = {m_{dd}}.C{\% _{{H_2}S{O_4}}} = 150.19,6\%  = 29,4{\text{ gam}}\)

\( \to {n_{{H_2}S{O_4}}} = \frac{{{m_{{H_2}S{O_4}}}}}{{{M_{{H_2}S{O_4}}}}} = \frac{{29,4}}{{98}} = 0,3{\text{ mol}}\)

Theo tỉ lệ phản ứng:

\({n_{{H_2}}} = {n_{{H_2}S{O_4}}} = 0,3{\text{ mol}}\)

\( \to {V_{{H_2}}} = 0,3.22,4 = 6,72{\text{ lít}}\)

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

\( \to {m_{Al}} = {n_{Al}}.{M_{Al}} = 0,2.27 = 5,4{\text{ gam}}\)

Đây nhaaa