1)

a) Tác dụng với nước

\(N{a_2}O + {H_2}O\xrightarrow{{}}2NaOH\)

\(S{O_2} + {H_2}O \rightleftharpoons {H_2}S{O_3}\)

b)

Tác dụng với \(HCl\)

\(N{a_2}O + 2HCl\xrightarrow{{}}2NaCl + {H_2}O\)

\(F{e_2}{O_3} + 6HCl\xrightarrow{{}}2FeC{l_3} + 3{H_2}O\)

c)

\(S{O_2} + 2NaOH\xrightarrow{{}}N{a_2}S{O_3} + {H_2}O\)

2)

Các phản ứng xảy ra:

\(Mg{(OH)_2} + {H_2}S{O_4}\xrightarrow{{}}MgS{O_4} + 2{H_2}O\)

\(Fe + {H_2}S{O_4}\xrightarrow{{}}FeS{O_4} + {H_2}\)

\(BaC{l_2} + {H_2}S{O_4}\xrightarrow{{}}BaS{O_4} + 2HCl\)

\(CaO + {H_2}S{O_4}\xrightarrow{{}}CaS{O_4} + {H_2}O\)

3)

Phản ứng xảy ra:

\(Mg + {H_2}S{O_4}\xrightarrow{{}}MgS{O_4} + {H_2}\)

\(MgO+ {H_2}S{O_4}\xrightarrow{{}}MgS{O_4} + {H_2}O\)

\({n_{{H_2}}} = \frac{{2,24}}{{22,4}} = 0,1{\text{ mol = }}{{\text{n}}_{Mg}}\)

\( \to {m_{Mg}} = 0,1.24 = 2,4{\text{ gam}} \to {m_{MgO}} = 10{\text{ gam}}\)

\( \to \% {m_{Mg}} = \frac{{2,4}}{{12,4}} = 19,35\%  \to \% {m_{MgO}} = 80,65\% \)

\({n_{MgO}} = \frac{{10}}{{40}} = 0,25{\text{ mol}}\)

\( \to {n_{{H_2}S{O_4}}} = {n_{Mg}} + {m_{MgO}} = 0,25 + 0,1 = 0,35{\text{ mol}}\)

\({m_{{H_2}S{O_4}}} = 0,35.98 = 34,3{\text{ gam}}\)

\( \to {m_{dd{\text{ }}{{\text{H}}_2}S{O_4}}} = \frac{{34,3}}{{9,8\% }} = 350{\text{ gam}}\)

BTKL:

\({m_X} + {m_{dd{{\text{H}}_2}S{O_4}}} = {m_{dd}} + {m_{{H_2}}}\)

\( \to {m_{dd}} = 12,4 + 350 - 0,1.2 = 362,2{\text{ gam}}\)

\({n_{MgS{O_4}}} = {n_{Mg}} + {n_{MgO}} = 0,1 + 0,25 = 0,35{\text{ mol}}\)

\( \to {m_{MgS{O_4}}} = 0,35.(24 + 96) = 42{\text{ gam}}\)

\( \to C{\% _{MgS{O_4}}} = \frac{{42}}{{362,2}} = 11,6\% \)