Đáp án:
\({m_{KCl{O_3}}} = 4,08333{\text{ gam}}\)
\({V_{S{O_2}}} = 1,12{\text{ lít}}\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(2KCl{O_3}\xrightarrow{{{t^o}}}2KCl + 3{O_2}\)
Ta có:
\({n_{{O_2}}} = \frac{{1,12}}{{22,4}} = 0,05{\text{ mol}}\)
\( \to {n_{KCl{O_3}}} = \frac{2}{3}{n_{{O_2}}} = \frac{{0,1}}{3}{\text{ mol}}\)
\( \to {m_{KCl{O_3}}} = \frac{{0,1}}{3}.(39 + 35,5.16.3) = 4,08333{\text{ gam}}\)
\(S + {O_2}\xrightarrow{{{t^o}}}S{O_2}\)
\( \to {n_{S{O_2}}} = {n_{{O_2}}} = 0,05{\text{ mol}}\)
\( \to {V_{S{O_2}}} = 0,05.22,4 = 1,12{\text{ lít}}\)