Đáp án:
\({m_{Ca{{(HS{O_3})}_2}}} = 5,656{\text{ gam}}\)
\({m_{CaS{O_3}}} = 10,08{\text{ gam}}\)
\({C_{M{\text{ Ca(HS}}{{\text{O}}_3}{)_2}}} = 0,07M\)
Giải thích các bước giải:
Ta có:
\({n_{S{O_2}}} = \frac{{3,136}}{{22,4}} = 0,14{\text{ mol;}}{{\text{n}}_{Ca{{(OH)}_2}}} = 0,4.0,28 = 0,112{\text{ mol}}\)
Ta có:
\(\frac{{{n_{S{O_2}}}}}{{{n_{Ca{{(OH)}_2}}}}} = \frac{{0,14}}{{0,112}} = 1,25\)
Vậy phản ứng tạo 2 muối.
\(Ca{(OH)_2} + S{O_2}\xrightarrow{{}}CaS{O_3} + {H_2}O\)
\(Ca{(OH)_2} + 2S{O_2}\xrightarrow{{}}Ca{(HS{O_3})_2}\)
\( \to {n_{Ca{{(HS{O_3})}_2}}} = 0,14 - 0,112 = 0,028{\text{ mol}}\)
\({n_{CaS{O_3}}} = 0,112 - 0,028 = 0,084{\text{ mol}}\)
\( \to {m_{Ca{{(HS{O_3})}_2}}} = 0,028.(40 + 81.2) = 5,656{\text{ gam}}\)
\({m_{CaS{O_3}}} = 0,084.(40 + 80) = 10,08{\text{ gam}}\)
\({C_{M{\text{ Ca(HS}}{{\text{O}}_3}{)_2}}} = \frac{{0,028}}{{0,4}} = 0,07M\)
