a) PTHH: Fe + 2HCl → FeCl₂ + H₂ (1)

FeCl₂ + 2AgNO₃ → Fe(NO₃)₂ + 2AgCl (2)

b)

\({n_} = \frac{V}{{22,4}} = \frac{{6,72}}{{22,4}} = 0,3mol\)

Theo phương trình (1)

\({n_} = {n_{Fe}} = 0,3mol\)

=> \(a = {m_{Fe}} = 0,3 \times 56 = 16,8g\)

Theo phương trình (2)

\({n_{AgCl}} = 2{n_{FeC{l_2}}} = 0,6mol = > b = {m_{AgCl}} = 0,6 \times 143,5 = 86,1g\)

c)

\({C_M} = \frac{n}{V} = \frac{{0,6}}{{0,15}} = 4M\)

\(C{\% _{HCl}} = \frac{{{m_{ct}}}}{{{m_{dd}}}} \times 100 = \frac{{0,6 \times 36,5}}{{1,2 \times 150}} \times 100 = 12,17\% \)