$\text{n$_{O2}$=8,4:22,4=0,375(mol)}$

$\text{n$_{Fe3O4}$=34,8:232=0,15(mol)}$

$\text{PTHH:}$

$\text{3Fe+2O$_{2}$$\xrightarrow[]{t^o}$Fe$_{3}$O$_{4}$}$

$\text{0,45←0,3  ←0,15(mol)}$

$\text{a,Từ PTHH ta suy ra : n$_{Fe}$=0,45(mol)}$

$\text{⇒m$_{Fe}$=0,45.56=25,2(g)}$

$\text{b,Từ PTHH ta suy ra : n$_{O2 phản ứng}$=0,3(mol)}$

$\text{⇒n$_{O2 dư}$=0,375-0,3=0,075(mol)}$

$\text{⇒V$_{O2 dư}$=0,075.22,4=1,68(lít)}$

$\text{c,PTHH:}$

$\text{2KMnO$_{4}$$\xrightarrow[]{t^o}$O$_{2}$+K$_{2}$MnO$_{4}$+MnO$_{2}$}$

$\text{0,75    ←0,375(mol)}$

$\text{⇒m$_{KMnO4}$=0,75.158=118,5(g)}$

$\text{$n_{O_2}$ = $\dfrac{V}{22,4}$ = $\dfrac{8,4}{22,4}$ = 0,375 mol}$

$\rightarrow$ $\text{$m_{O_2}$ = n. M = 0,375. 32 = 12 gam}$

a, $\text{Đặt CTHH của Oxit sắt từ là $Fe_xO_y$ }$

Theo định luật bảo toàn khối lượng:

$\text{$m_{Fe}$ + $m_{O_2}$ = $m_{Fe_xO_y}$ }$

$\text{$m_{Fe}$ = $m_{Fe_xO_y}$ - $m_{O_2}$ }$

$\text{$m_{Fe}$ = 34,8 - 12 = 22,8 gam}$

a, PTHH: $\text{$2KMnO_{4}$ $\xrightarrow{t^o}$ $K_{2}$$MnO_{4}$ + $O_2$ + $MnO_{2}$ }$

Theo PTHH $\rightarrow$ $\text{$n_{KMnO_4}$ = $2n_{O_2}$ = 0,375. 2 = 0,75 mol}$

$\rightarrow$ $\text{$m_{KMnO_4}$ = n. M = 0,75. 158 = 118,5 gam}$

$\text{#Kudo}$