Đáp án:
\({m_{Fe}} = 16,8{\text{ gam; }}{{\text{V}}_{{O_2}}} = 4,48{\text{ lít}}\)
\({{\text{m}}_{KMn{O_4}}} = 63,2{\text{ gam}}\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(3Fe + 2{O_2}\xrightarrow{{{t^o}}}F{e_3}{O_4}\)
Ta có: \({n_{F{e_3}{O_4}}} = \frac{{23,2}}{{56.3 + 16.4}} = 0,1{\text{ mol}}\)
Ta có: \({n_{Fe}} = 3{n_{F{e_3}{O_4}}} = 0,3{\text{ mol}} \to {{\text{m}}_{Fe}} = 0,3.56 = 16,8{\text{ gam}}\)
\({n_{{O_2}}} = 2{n_{F{e_3}{O_4}}} = 0,2{\text{ mol}} \to {{\text{V}}_{{O_2}}} = 0,2.22,4 = 4,48\;{\text{lít}}\)
\(2KMn{O_4}\xrightarrow{{}}{K_2}Mn{O_4} + Mn{O_2} + {O_2}\)
\(\to {n_{KMn{O_4}}} = 2{n_{{O_2}}} = 0,4{\text{ mol}} \to {{\text{m}}_{KMn{O_4}}} = 0,4.(39 + 55 + 16.4) = 63,2{\text{ gam}}\)