1)
a)
\(N{a_2}S{O_4} + BaC{l_2}\xrightarrow{{}}BaS{O_4} + 2NaCl\)
\(NaCl + {H_2}S{O_4}\xrightarrow{{{t^o}}}NaHS{O_4} + HCl\)
\(Mn{O_2} + 4HCl\xrightarrow{{}}MnC{l_2} + C{l_2} + 2{H_2}O\)
\(C{l_2} + 2NaOH\xrightarrow{{}}NaCl + NaClO + {H_2}O\)
b)
\(S + {O_2}\xrightarrow{{{t^o}}}S{O_2}\)
\(S{O_2} + 2{H_2}S\xrightarrow{{{t^o}}}3S + 2{H_2}O\)
\(2S{O_2} + {O_2}\xrightarrow{{{V_2}{O_5},{t^o}}}2S{O_3}\)
\(S{O_3} + {H_2}O\xrightarrow{{}}{H_2}S{O_4}\)
2)
Phản ứng xảy ra:
\(Mg + {H_2}S{O_4}\xrightarrow{{}}MgS{O_4} + {H_2}\)
\(2Al + 3{H_2}S{O_4}\xrightarrow{{}}A{l_2}{(S{O_4})_3} + 3{H_2}\)
Gọi số mol Mg và Al trong hỗn hợp ban đầu lần lượt là x, y.
\( \to 24x + 27y = 15,6{\text{ gam}}\)
\({n_{{H_2}}} = x + 1,5y = \frac{{17,92}}{{22,4}} = 0,8{\text{ mol}}\)
Giải được: x=0,2; y=0,4.
\( \to {m_{Mg}} = 0,2.24 = 4,8{\text{ gam;}}{{\text{m}}_{Al}} = 0,4.27 = 10,8{\text{ gam}}\)
\({n_{{H_2}S{O_4}}} = {n_{{H_2}}} = 0,8{\text{ mol}} \to {{\text{V}}_{dd{\text{ }}{{\text{H}}_2}S{O_4}}} = \frac{{0,8}}{2} = 0,4{\text{ lít}}\)
