1)

Phản ứng xảy ra:

\(F{\text{e}} + 2HCl\xrightarrow{{}}F{\text{e}}C{l_2} + {H_2}\)

Ta có:

\({n_{F{\text{e}}}} = \frac{{11,2}}{{56}} = 0,2{\text{ mol;}}{{\text{n}}_{HCl}} = \frac{{7,3}}{{36,5}} = 0,2{\text{ mol}}\)

Suy ra:

\({n_{F{\text{e}}}} > \frac{1}{2}{n_{HCl}}\) nên \(Fe\) dư

\( \to {n_{F{\text{e dư}}}} = 0,2 - \frac{{0,2}}{2} = 0,1{\text{ mol}}\)

\( \to {m_{F{\text{e dư}}}} = 0,1.56 = 5,6{\text{ gam}}\)

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

2)

Phản ứng xảy ra:

\(F{\text{e}} + 2HCl\xrightarrow{{}}F{\text{e}}C{l_2} + {H_2}\)

Ta có:

\({n_{F{\text{e}}}} = \frac{{5,6}}{{56}} = 0,1{\text{ mol;}}{{\text{n}}_{HCl}} = \frac{{3,65}}{{36,5}} = 0,1{\text{ mol}}\)

Suy ra:

\({n_{F{\text{e}}}} > \frac{1}{2}{n_{HCl}}\) nên \(Fe\) dư

\( \to {n_{F{\text{e dư}}}} = 0,1 - \frac{{0,1}}{2} = 0,05{\text{ mol}}\)

\( \to {m_{F{\text{e dư}}}} = 0,05.56 = 2,8{\text{ gam}}\)

\({n_{{H_2}}} = \frac{1}{2}{n_{HCl}} = 0,05{\text{ mol}} \to {{\text{V}}_{{H_2}}} = 0,05.22,4 = 1,12{\text{ lít}}\)

3)

Phản ứng xảy ra:

\(F{\text{e}} + {H_2}S{O_4}\xrightarrow{{}}F{\text{e}}S{O_4} + {H_2}\)

Ta có:

\({n_{F{\text{e}}}} = \frac{{11,2}}{{56}} = 0,2{\text{ mol;}}{{\text{n}}_{{H_2}S{O_4}}} = \frac{{29,4}}{{98}} = 0,3{\text{ mol}}\)

Suy ra:

\({n_{F{\text{e}}}} < {n_{{H_2}S{O_4}}}\) nên \(H_2SO_4\) dư

\( \to {n_{{H_2}S{O_4}{\text{ dư}}}} = 0,3 - 0,2 = 0,1{\text{ mol}}\)

\( \to {m_{{H_2}S{O_4}{\text{ dư}}}} = 0,1.98 = 9,8{\text{ gam}}\)