Đáp án:
\( \to m = 27,9{\text{ gam}}\)
Giải thích các bước giải:
Ta có:
\({n_{S{O_2}}} = \frac{{3,36}}{{22,4}} = 0,15{\text{ mol}}\)
\({n_{KOH}} = 0,2.1,875 = 0,375{\text{ mol}}\)
\( \to T = \frac{{{n_{KOH}}}}{{{n_{S{O_2}}}}} = \frac{{0,375}}{{0,15}} > 2\)
Vậy \(KOH\) dư
\( \to {n_{{K_2}S{O_3}}} = {n_{S{O_2}}} = 0,15{\text{ mol}}\)
\({n_{KOH{\text{ dư}}}} = 0,375 - 0,15.2 = 0,075{\text{ mol}}\)
\( \to m = {m_{{K_2}S{O_3}}} + {m_{KOH{\text{ dư}}}} \)
\(= 0,15.(39.2 + 32 + 16.3) + 0,075.56 = 27,9{\text{ gam}}\)