\(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\) ; \(n_{O_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

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

0,2  <  0,25                        ( mol )

0,2      \(\dfrac{2}{15}\)                 \(\dfrac{1}{15}\)   ( mol )

`->` Chất dư là O₂

\(m_{O_2\left(dư\right)}=\left(0,25-\dfrac{2}{15}\right).32=3,73\left(g\right)\)

\(V_{kk}=V_{O_2}.5=\dfrac{2}{15}.22,4.5=14,93\left(l\right)\)

\(m_{bôt.sắt}=\dfrac{11,2.100}{100-12}=12,72\left(g\right)\)

\(n_{Al}=\dfrac{6,75}{27}=0,25mol\)

\(n_{O_2}=\dfrac{3,36}{22,4}=0,15mol\)

\(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)

0,25   0,15     0

0,2     0,15     0,1

0,05   0          0,1

\(m_{dư}=m_{Aldư}=0,05\cdot27=1,35g\)

\(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\uparrow\)

0,3                                                 0,15

\(m_{KMnO_4}=0,3\cdot158=47,4g\)

 \(4P+5O_2\underrightarrow{t^o}2P_2O_5\)

\(nO_2=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

\(\Rightarrow nP_2O_5=\dfrac{2}{5}nO_2=\dfrac{2}{5}.0,1=0,04\left(mol\right)\)

\(\Rightarrow mP_2O_5=0,04.\left(31.2+16.5\right)=5,68\left(g\right)\)

c/

pthh: \(2KMnO_4\rightarrow K_2MnO_4+MnO_2+O_2\uparrow\)

\(nO_2=0,1\left(mol\right)\)

\(\Rightarrow nKMnO_4=\dfrac{2}{1}.nO_2=\dfrac{2}{1}.0,1=0,2\left(mol\right)\)

\(\Rightarrow mKMnO_4=0,2.\left(39+55+16.4\right)=31,6\left(g\right)\)

\(n_{O_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)

PTHH: 2KMnO₄ --t^o--> K₂MnO₄ + MnO₂ + O₂

               1<------------------------------------0,5

=> \(m_{KMnO_4\left(pthh\right)}=1.158=158\left(g\right)\)

=> \(m_{KMnO_4\left(tt\right)}=\dfrac{158.100}{80}=197,5\left(g\right)\)

\(n_{Fe}=\dfrac{8,4}{56}=0,15\left(mol\right)\\ pthh:3Fe+2O_2\underrightarrow{t^o}Fe_3O_{\text{4}}\) 
            0,15     0,1       0,05 
\(m_{Fe_2O_4}=0,05.232=11,6\left(g\right)\\ V_{O_2}=0,1.11,4=2,24\left(l\right)\\ pthh:2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\uparrow\) 
                0,2                                                0,1 
\(m_{KMnO_4}=0,2.158=31,6\left(g\right)\)

\(n_{Fe}=\dfrac{8,4}{56}=0,15\left(mol\right)\\ pthh:3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\) 
            0,15    0,1      0,05 
\(m_{Fe_3O_{\text{ 4}}}=0,05.232=11,6\left(g\right)\\ V_{O_2}=0,1.22,4=2,24\left(l\right)\\ pthh:2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\) 
              0,1                                               0,05 
\(m_{KMnO_4}=0,1.158=15,8\left(g\right)\)

a, PT: \(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)

b, Ta có: \(n_{O_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

Theo PT: \(n_{Fe_3O_4}=\dfrac{1}{2}n_{O_2}=0,1\left(mol\right)\)

\(\Rightarrow m_{Fe_3O_4}=0,1.232=23,2\left(g\right)\)

c, PT: \(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)

Có: O₂ hao hụt 40% → H% = 100 - 40 = 60%

Theo PT: \(n_{KMnO_4}=2n_{O_2}=0,4\left(mol\right)\)

\(\Rightarrow n_{KMnO_4\left(TT\right)}=\dfrac{0,4}{60\%}=\dfrac{2}{3}\left(mol\right)\)

\(\Rightarrow m_{KMnO_4}=\dfrac{2}{3}.158\approx105,3\left(g\right)\)

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