Đáp án:

\(m = 21,4375{\text{ gam ; }}{{\text{m}}_1} = 18,75{\text{ gam}}\)

Giải thích các bước giải:

 Phản ứng xảy ra:

\(2KCl{O_3}\xrightarrow{{Mn{O_2},{t^o}}}2KCl + 3{O_2}\)

\(4Al + 3{O_2}\xrightarrow{{{t^o}}}2A{l_2}{O_3}\)

\(2Mg + {O_2}\xrightarrow{{{t^o}}}2MgO\)

Gọi số mol Al là x suy ra Mg là 0,6x

\( \to 27x + 24.0,6x = 10,35 \to x = 0,25{\text{ mol}}\)

\( \to {n_{A{l_2}{O_3}}} = \frac{1}{2}{n_{Al}} = 0,125{\text{ mol; }}{{\text{n}}_{MgO}} = {n_{MgO}} = 0,15{\text{ mol}}\)

\( \to {m_1} = {m_{A{l_2}{O_3}}} + {m_{MgO}} = 0,125.102 + 0,15.40 = 18,75{\text{ gam}}\)

\({n_{{O_2}}} = \frac{3}{2}{n_{A{l_2}{O_3}}} + \frac{1}{2}{n_{MgO}} = \frac{3}{2}.0,125 + \frac{1}{2}.0,15 = 0,2625{\text{ mol}}\)

\( \to {n_{KCl{O_3}}} = \frac{2}{3}{n_{{O_2}}} = 0,175{\text{ mol}} \to {{\text{m}}_{KCl{O_3}}} = 0,175.(39 + 35,5 + 16.3) = m = 21,4375{\text{ gam}}\)