Đáp án:
\({m_{rắn}} = 14,4{\text{ gam}}\)
\({m_{KCl{O_3}}} = 8,167{\text{ gam; }}{{\text{m}}_{KMn{O_4}}} = 42,133{\text{ gam}}\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(3Fe + 2{O_2}\xrightarrow{{{t^o}}}F{e_3}{O_4}\)
Ta có: \({n_{Fe}} = \frac{{11,2}}{{56}} = 0,2{\text{ mol ;}}{{\text{n}}_{{O_2}}} = \frac{{2,24}}{{22,4}} = 0,1{\text{ mol}} \to {{\text{n}}_{Fe}} > \frac{3}{2}{n_{{O_2}}}\) nên Fe dư.
BTKL: \({m_{rắn}} = {m_{Fe}} + {m_{{O_2}}} = 11,2 + 0,1.32 = 14,4{\text{ gam}}\)
\(2KCl{O_3}\xrightarrow{{}}2KCl + 3{O_2}\)
Ta có: \({n_{KCl{O_3}}} = \frac{2}{3}{n_{{O_2}}} = \frac{{0,2}}{3}{\text{ mol}} \to {{\text{m}}_{KCl{O_3}}} = \frac{{0,2}}{3}(39 + 35,5 + 16.3) = 8,167{\text{ gam}}\)
\(2KMn{O_4}\xrightarrow{{}}{K_2}Mn{O_4} + Mn{O_2} + {O_2}\)
\({n_{{O_2}{\text{ cần dùng}}}} = \frac{2}{3}{n_{Fe}} = 0,2.\frac{2}{3} = \frac{{0,4}}{3}{\text{ mol}} \to {{\text{n}}_{KMn{O_4}}} = 2{n_{{O_2}}} = \frac{{0,8}}{3}{\text{ mol}} \to {{\text{m}}_{KMn{O_4}}} = \frac{{0,8}}{3}(39 + 55 + 16.4) = 42,133{\text{ gam}}\)
