1)

Phản ứng xảy ra:

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

Ta có:

\({n_{{H_2}}} = \frac{{2,8}}{{22,4}} = 0,125{\text{ mol = }}{{\text{n}}_{Fe}}\)

\( \to {m_{Fe}} = 0,125.56 = 7{\text{ gam}}\)

\( \to \% {m_{Fe}} = \frac{7}{{10}} = 70\% \)

2)

Phản ứng xảy ra:

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

Ta có:

\({n_{{H_2}}} = \frac{{1,12}}{{22,4}} = 0,05{\text{ mol = }}{{\text{n}}_{Fe}}\)

\( \to {m_{Fe}} = 0,05.56 = 2,8{\text{ gam}}\)

Rắn không tan là \(Ag\)

\( \to {m_A} = {m_{Ag}} = 4 - 2,8 = 1,2{\text{ gam}}\)

3)

Phản ứng xảy ra:

\(2KMn{O_4} + 16HCl\xrightarrow{{}}2KCl + 2MnC{l_2} + 5C{l_2} + 8{H_2}O\)

Ta có:

\({n_{HCl}} = \frac{{21,9}}{{36,5}} = 0,6{\text{ mol;}}{{\text{n}}_{KMn{O_4}}} = \frac{{15,8}}{{39 + 55 + 16.4}} = 0,1{\text{ mol}}\)

Vì \({n_{HCl}} < 8{n_{KMn{O_4}}}\)

Nên \(KMnO_4\) dư

\( \to {n_{C{l_2}}} = \frac{5}{{16}}{n_{HCl}} = \frac{5}{{16}}.0,6 = 0,1875{\text{ mol}}\)

\( \to {V_{C{l_2}}} = 0,1875.22,4 = 4,2{\text{ lít}}\)

4)

Phản ứng xảy ra:

\(2Al + 3{X_2}\xrightarrow{{{t^o}}}2Al{X_3}\)

Ta có:

\({n_{Al}} = \frac{{2,7}}{{27}} = 0,1{\text{ mol}}\)

\( \to {n_{{X_2}}} = \frac{3}{2}{n_{Al}} = 0,15{\text{ mol}}\)

\( \to {M_{{X_2}}} = \frac{{10,65}}{{0,15}} = 71 \to {M_X} = 35,5 \to X:Cl\)

Vậy halogen là \(Cl_2\)