Đáp án:
\( {m_{MgO}} = 5,67{\text{ gam}}\)
\({V_{{O_2}}} = 1,904{\text{ lít}}\)
\( {m_{KCl{O_3}}} = 7,713{\text{ gam}}\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(2Mg + {O_2}\xrightarrow{{{t^o}}}2MgO\)
Ta có:
\({n_{Mg}} = \frac{{3,4}}{{24}} = {n_{MgO}}\)
\( \to {m_{MgO}} = \frac{{3,4}}{{24}}.(24 + 16) = 5,67{\text{ gam}}\)
\({n_{{O_2}{\text{ phản ứng}}}} = \frac{1}{2}{n_{Mg}} = \frac{{1,7}}{{24}}\)
\( \to {n_{{O_2}}} = \frac{{1,7}}{{24}}.120\% = 0,085{\text{ mol}}\)
\( \to {V_{{O_2}}} = 0,085.22,4 = 1,904{\text{ lít}}\)
\(2KCl{O_3}\xrightarrow{{{t^o}}}2KCl + 3{O_2}\)
Ta có:
\({n_{KCl{O_3}{\text{ lt}}}} = \frac{2}{3}{n_{{O_2}}} = \frac{{0,17}}{3}\)
Vì hao hụt \(10\%\)
\( \to {n_{KCl{O_3}}} = \frac{{\frac{{0,17}}{3}}}{{90\% }} = \frac{{17}}{{270}} \to {m_{KCl{O_3}}} = \frac{{0,17}}{{270}}.(39 + 35,5 + 16.3) = 7,713{\text{ gam}}\)