Đáp án:
\( {m_{CaS{O_3}}} = 6{\text{ gam}}\)
Giải thích các bước giải:
Ta có:
\({n_{S{O_2}}} = \frac{{2,24}}{{22,4}} = 0,1{\text{ mol}}\)
\({n_{NaOH}} = \frac{{12}}{{40}} = 0,3{\text{ mol}}\)
\( \to \frac{{{n_{NaOH}}}}{{{n_{S{O_2}}}}} = \frac{{0,3}}{{0,1}} > 2\)
Do vậy \(NaOH\) dư
Phản ứng xảy ra:
\(2NaOH + S{O_2}\xrightarrow{{}}N{a_2}S{O_3} + {H_2}O\)
\( \to {n_{NaOH{\text{ dư}}}} = 0,3 - 0,1.2 = 0,1{\text{ mol}}\)
\( \to {m_{NaOH{\text{ dư}}}} = 0,1.40 = 4{\text{ gam}}\)
\({n_{N{a_2}S{O_3}}} = {n_{S{O_2}}} = 0,1{\text{ mol}}\)
\( \to {n_{N{a_2}S{O_3}}} = 0,1.(23.2 + 32 + 16.3) = 12,6{\text{ gam}}\)
Dẫn khí tác dụng với \(Ca(OH)_2\)
Ta có:
\({n_{Ca{{(OH)}_2}}} = 0,075.0,1 = 0,075{\text{ mol}}\)
\( \to 1 < \frac{{{n_{C{O_2}}}}}{{{n_{Ca{{(OH)}_2}}}}} < 2\) nên tạo hỗn hợp muối.
\(Ca{(OH)_2} + S{O_2}\xrightarrow{{}}CaS{O_3} + {H_2}O\)
\(Ca{(OH)_2} + 2S{O_2}\xrightarrow{{}}Ca{(HS{O_3})_2}\)
\( \to {n_{Ca{{(HS{O_3})}_2}}} = {n_{S{O_2}}} - {n_{Ca{{(OH)}_2}}} = 0,1 - 0,075 = 0,025{\text{ mol}}\)
\( \to {n_{CaS{O_3}}} = 0,075 - 0,025 = 0,05{\text{ mol}}\)
\( \to {m_{CaS{O_3}}} = 0,05.(40 + 32 + 16.3) = 6{\text{ gam}}\)
