a) $3Fe +2 O_2 \xrightarrow{t^o} Fe_3O_4$

b) $n_{O_2} = \dfrac{4,48}{22,4} = 0,2(mol)$
Theo PTHH : $n_{Fe} = \dfrac{3}{2}n_{O_2} = 0,3(mol)$
$m_{Fe} = 0,3.56 = 16,8(gam)$

c) $n_{Fe_3O_4} = \dfrac{1}{2}n_{O_2} = 0,1(mol)$
$m_{Fe_3O_4} = 0,1.232 = 23,2(gam)$

a) Chất tham gia: Sắt (Fe), Oxi (O₂)

Sản phẩm: Sắt từ (Fe₃O₄)

b) Theo ĐLBTKL

 \(m_{Fe}+m_{O_2}=m_{Fe_3O_4}\) (1)

c) \(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\); \(n_{Fe_3O_4}=\dfrac{46,4}{232}=0,2\left(mol\right)\)

PTHH: 3Fe + 2O₂ --t^o--> Fe₃O₄

______0,2----------------->\(\dfrac{0,2}{3}\) ________(mol)

=> vô lí ...

a)

\(3Fe + 2O_2 \xrightarrow{t^o} Fe_3O_4\)

b)

Ta có :

\(n_{Fe} = \dfrac{8,4}{56} = 0,15(mol)\\ n_{O_2} = \dfrac{96}{32} = 3(mol)\)

Ta thấy : \(\dfrac{n_{Fe}}{3} = 0,05 < \dfrac{n_{O_2}}{2} = 1,5\) do đó O₂ dư.

Theo PTHH :

\(n_{O_2\ pư} = \dfrac{2}{3}n_{Fe} = 0,1(mol)\\ \Rightarrow n_{O_2\ dư} = 3 - 0,1 = 2,9(mol)\\ \Rightarrow m_{O_2\ dư} = 92,8(gam)\)

c)

\(n_{Fe_3O_4} = \dfrac{1}{3}n_{Fe} = 0,05(mol)\\ \Rightarrow m_{Fe_3O_4} = 0,05.232 = 11,6(gam)\)

\(a)PTHH:FeCl_3+2O_2\xrightarrow[]{t^o}Fe_3O_4\)

mol                 1          2              1

mol

\(b)\)Số mol \(FeCl_3\) là: \(n_{FeCl_3}=\dfrac{m_{FeCl_3}}{M_{FeCl_3}}=\dfrac{8,4}{162,5}=0,052\left(mol\right)\)

Số mol \(O_2\) là: \(n_{O_2}=\dfrac{m_{O_2}}{M_{O_2}}=\dfrac{96}{32}=3\left(mol\right)\)

Lập tỉ lệ: \(\dfrac{1}{0,052}>\dfrac{2}{3}\Rightarrow FeCl_3dư\)

Số mol \(FeCl_3\) phản ứng là:

Từ PTHH\(\Rightarrow\) \(n_{FeCl_3}=\dfrac{0,052\times3}{3}=0,035\left(mol\right)\)

Số mol \(FeCl_3\) dư là: \(n_{FeCl_3dư}=n_{FeCl_3đầu}-n_{FeCl_3p/ứng}=0,052-0,035=0,018\left(mol\right)\)

Khối lượng \(FeCl_3\) dư là: \(m_{FeCl_3dư}=n_{FeCl_3dư}\times M_{FeCl_3}=0,018\times162,5=2,925\left(g\right)\)

\(n_{Fe_3O_4}=\dfrac{m_{Fe_3O_4}}{M_{Fe_3O_4}}=\dfrac{23,2}{232}=0,1mol\)

\(3Fe+2O_2\rightarrow\left(t^o\right)Fe_3O_4\)

0,3       0,2              0,1          ( mol )

\(m_{Fe}=n_{Fe}.M_{Fe}=0,3.56=16,8g\)

\(V_{O_2}=n_{O_2}.22,4=0,2.22,4=4,48l\)

\(V_{kk}=\dfrac{4,48.100}{20}=22,4l\)

\(2KMnO_4\rightarrow\left(t^o\right)K_2MnO_4+MnO_2+O_2\)

    0,4                                         0,2               ( mol )

\(n_{KMnO_4}=\dfrac{0,4}{85\%}=\dfrac{8}{17}mol\)

\(m_{KMnO_4}=n_{KMnO_4}.M_{KMnO_4}=\dfrac{8}{17}.158=74,3529g\)

3Fe+2O₂-to>Fe₃O₄

0,3------0,2-------0,1mol

n Fe3O4=\(\dfrac{23,2}{232}\)=0,1 mol

->m Fe=0,3.56=16,8g

b)

VO2=0,2.22,4=4,48l

=>Vkk=4,48.5=22,4l

c)2KMnO₄-to>K₂MnO₄+MnO₂+O₂

   0,4-------------------------------------0,2

=>m KMnO4=0,4.158.\(\dfrac{100}{85}\)=74,35g

nFe = 11.2/56=0.2 (mol) 

3Fe + 2O2 -to-> Fe3O4 

0.2____2/15____1/15

VO2 = 2/15 * 22.4 = 2.9867 (l) 

mFe3O4 = 1/15 * 232  = 15.47 (g) 

ta có pthh: 3Fe + 2O₂ → Fe₃O₄

Ta có nFe=\(\dfrac{m}{M}\)=\(\dfrac{11,2}{56}\)=0,2(mol)

          nO₂=2nFe=2*\(\dfrac{0,2}{3}\)=\(\dfrac{2}{15}\)(mol)

          VO₂=n*M=16*\(\dfrac{2}{15}\)=2,13(l)

          nFe₃O₄=\(\dfrac{0,2}{2}\)=0,1(mol)

          mFe₃O₄=\(\dfrac{0,1}{168+64}\)=23,2(g)

\(a)3Fe+2O_2\rightarrow Fe_3O_4\)

\(3mol\)   \(2mol\)        \(1mol\)

\(0,3mol\)  \(0,2mol\)   \(0,1mol\)

\(b)n_{Fe}=\dfrac{m}{M}=\dfrac{16,8}{56}=0,3\left(mol\right)\)

\(n_{O_2}=\dfrac{V}{22,4}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

\(\text{Ta thấy }O_2\text{ dư,}Fe\text{ phản ứng hết}\)

\(c)m_{Fe_3O_4}=n.M=0,1.232=23,2\left(g\right)\)

\(a,3Fe+2O_2\rightarrow\left(t^o\right)Fe_3O_4\\ n_{Fe}=\dfrac{16,8}{56}=0,3\left(kmol\right)\\ n_{O_2}=\dfrac{2}{3}.0,3=0,2\left(kmol\right)\\ V_{O_2\left(\text{đ}ktc\right)}=0,2.1000.22,4=4480\left(l\right)\\ n_{Fe_3O_4}=\dfrac{1}{3}.0.3=0,1\left(kmol\right)\\ m_{Fe_3O_4}=232.0,1=23,2\left(kg\right)\)