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: \({n_{Fe}} = 3{n_{F{e_3}{O_4}}} = 0,3{\text{ mol}} \to {{\text{m}}_{Fe}} = 0,3.56 = 16,8{\text{ gam}}\)

\({n_{{O_2}}} = 2{n_{F{e_3}{O_4}}} = 0,2{\text{ mol}} \to {{\text{V}}_{{O_2}}} = 0,2.22,4 = 4,48\;{\text{lít}}\)

\(2KMn{O_4}\xrightarrow{{}}{K_2}Mn{O_4} + Mn{O_2} + {O_2}\)

\(\to {n_{KMn{O_4}}} = 2{n_{{O_2}}} = 0,4{\text{ mol}} \to {{\text{m}}_{KMn{O_4}}} = 0,4.(39 + 55 + 16.4) = 63,2{\text{ gam}}\)

$n_{Fe3O4}=23,2/232=0,1mol$

$a/$

$3Fe+2O2\overset{t^o}{\rightarrow}Fe3O4$

Theo pt:

$n_{Fe}=3.n_{Fe3O4}=0,1.3=0,3mol$

$⇒m_{Fe}=0,3.56=16,8g$

$b/$

Theo pt:

$n_{O2}=2.n_{Fe3O4}=2.0,1=0,2mol$

$⇒V_{O2}=0,2.22,4=4,48l$

$c/$

$2KMnO4\overset{t^o}{\rightarrow}K2MnO4+MnO2+O2$

$n_{KMnO4}=2.n_{O2}=2.0,2=0,4mol$

$⇒m_{KMnO4}=0,4.158=63,2g$