Đáp án:
\({m_{KCl{O_3}}} = 24,255{\text{ gam}}\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(2KCl{O_3}\xrightarrow{{{t^o}}}2KCl + 3{O_2}\)
\(C + {O_2}\xrightarrow{{{t^o}}}C{O_2}\)
Ta có:
\({m_C} = 3,96.(100\% - 10\% ) = 3,564{\text{ gam}}\)
\( \to {n_C} = \frac{{3,564}}{{12}} = 0,297{\text{ mol = }}{{\text{n}}_{{O_2}}}\)
\( \to {n_{KCl{O_3}}} = \frac{2}{3}{n_{{O_2}}} = \frac{2}{3}.0,297 = 0,198{\text{ mol}}\)
\( \to {m_{KCl{O_3}}} = 0,198.(39 + 35,5 + 16.3) = 24,255{\text{ gam}}\)
