Ta có:
\({n_{S{O_2}}} = \dfrac{{2,24}}{{22,4}} = 0,1{\text{ mol;}}{{\text{n}}_{Ca{{(OH)}_2}}} = 0,1.0,8 = 0,08{\text{ mol}}\)
\( \to \dfrac{{{n_{S{O_2}}}}}{{{n_{Ca{{(OH)}_2}}}}} = \dfrac{{0,1}}{{0,08}} = 1,25\)
Vậy phản ứng tạo 2 muối
\(Ca{(OH)_2} + S{O_2}\xrightarrow{{}}Ca{S{O}}_{3} + {H_2}O\)
\(Ca{(OH)_2} + 2S{O_2}\xrightarrow{{}}Ca{(HS{O_3})_2}\)
\( \to {n_{CaS{{\text{O}}_3}}} = 2{n_{Ca{{(OH)}_2}}} - {n_{S{O_2}}} = 0,08.2 - 0,1 = 0,06{\text{ mol}}\)
\( \to m = {m_{CaS{O_3}}} = 0,06.120 = 7,2{\text{ gam}}\)
