\(n_{Fe}=\dfrac{25,2}{56}=0,45\left(mol\right)\\ a,PTHH:3Fe+2O_2\rightarrow\left(t^o\right)Fe_3O_4\\ b,n_{O_2}=\dfrac{2}{3}.0,45=0,3\left(mol\right)\\ \Rightarrow V_{O_2\left(đktc\right)}=0,3.22,4=6,72\left(l\right)\\ c,2KClO_3\rightarrow\left(t^o\right)2KCl+3O_2\\ n_{KClO_3}=\dfrac{2}{3}.0,3=0,2\left(mol\right)\\ \Rightarrow m_{KClO_3}=122,5.0,2=24,5\left(g\right)\)

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

b, \(n_{Al}=\dfrac{8,1}{27}=0,3\left(mol\right)\)

Theo PT: \(n_{O_2}=\dfrac{3}{4}n_{Al}=0,225\left(mol\right)\Rightarrow V_{O_2}=0,225.22,4=5,04\left(l\right)\)

c, \(2KClO_3\underrightarrow{t^o}2KCl+3O_2\)

Theo PT: \(n_{KClO_3}=\dfrac{2}{3}n_{O_2}=0,15\left(mol\right)\Rightarrow m_{KClO_3}=0,15.122,5=18,375\left(g\right)\)

\(a,n_{Fe}=\dfrac{2,52}{56}=0,45\left(mol\right)\\ PTHH:3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\\ \left(mol\right)....0,45\rightarrow0,3...0,15\\ b,V_{O_2}=0,3.22,4=6,72\left(l\right)\\ c,PTHH:2KClO_3\underrightarrow{t^o}2KCl+3O_2\\ \left(mol\right).......0,2\leftarrow............0,3\\ m_{KClO_3}=0,2.122,5=24,5\left(g\right)\)

a) 3Fe + 2O2 -> Fe3O4 ( cần thêm đk nhiệt độ ở mũi tên )

b) nFe= 25,2/56=0,45(mol)

nO2= 2/3 . nFe = 2/3 . 0,45 = 0,3 ( mol )

-> VO2 = 0,3.22,4= 6,72(lít )

c) 2KClO3 -> 2KCl + 3O2  ( cần đk nhiệt độ )

nO2 = 0,3 ( mol ) 

nKClO3 = 2/3 . nO2 = 2/3 . 0,3 = 0,2 ( mol)

mKClO3= 0,2 . 122,5 = 24,5(g)

Câu 2: 

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

PTHH: \(2KMnO_4\xrightarrow[]{t^o}K_2MnO_4+MnO_2+O_2\)

               0,6<------------------------------------0,3

\(\Rightarrow m_{KMnO_4}=0,6.158=94,8\left(g\right)\)

Câu 3: 

\(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

             0,2<-----------------------0,2

=> m_Zn = 0,2.65 = 13 (g)

Câu 4:

\(n_{Fe}=\dfrac{22,4}{56}=0,4\left(mol\right)\)

PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

           0,4------------------------->0,4

             \(CuO+H_2\xrightarrow[]{t^o}Cu+H_2O\)

              0,4<---0,4

\(\Rightarrow m_{CuO}=0,4.80=32\left(g\right)\)

\(n_{Fe}=\dfrac{126}{56}=2,25\left(mol\right)\\ pthh:3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\) 
           2,25     1,5  
=> \(V_{O_2}=1,5.22,4=33,6\left(L\right)\) 
\(PTHH:2KClO_3\underrightarrow{t^o}2KCl+3O_2\) 
                1                             1,5 
=> \(m_{KClO3}=122,5\left(g\right)\)

a) 

\(n_{KClO_3}=\dfrac{24.5}{122.5}=0.2\left(mol\right)\)

\(2KClO_3\underrightarrow{^{^{t^0}}}2KCl+3O_2\)

\(n_{O_2}=\dfrac{3}{2}\cdot0.2=0.3\left(mol\right)\)

\(V_{O_2}=0.3\cdot22.4=6.72\left(l\right)\)

\(n_{O_2}=\dfrac{2.24}{22.4}=0.1\left(mol\right)\)

\(2KMnO_4\underrightarrow{^{^{t^0}}}K_2MnO_4+MnO_2+O_2\)

\(0.2...............................................0.1\)

\(n_{KMnO_4\left(bđ\right)}=\dfrac{0.2}{90\%}=\dfrac{2}{9}\left(mol\right)\)

\(m_{KMnO_4}=\dfrac{2}{9}\cdot158=35.11\left(g\right)\)

\(n_{Fe_3O_4}=\dfrac{3.48}{232}=0.015\left(mol\right)\)

\(3Fe+2O_2\underrightarrow{^{^{t^0}}}Fe_3O_4\)

\(0.045....0.03......0.015\)

\(m_{Fe}=0.045\cdot56=2.52\left(g\right)\)

\(V_{O_2}=0.03\cdot22.4=0.672\left(l\right)\)

\(2KClO_3\underrightarrow{^{^{t^0}}}2KCl+3O_2\)

\(0.02......................0.03\)

\(m_{KClO_3}=0.02\cdot122.5=2.45\left(g\right)\)

1. a. \(PTHH:2Mg+O_2\overset{t^o}{--->}2MgO\left(1\right)\)

b. Ta có: \(n_{Mg}=\dfrac{24}{24}=1\left(mol\right)\)

Theo PT₍₁₎: \(n_{O_2}=\dfrac{1}{2}.n_{Mg}=\dfrac{1}{2}.0,1=0,5\left(mol\right)\)

\(\Rightarrow V_{O_2}=0,5.22,4=11,2\left(lít\right)\)

c. \(PTHH:2KClO_3\xrightarrow[t^o]{MnO_2}2KCl+3O_2\left(2\right)\)

Theo PT₍₂₎: \(n_{KClO_3}=\dfrac{2}{3}.n_{O_2}=\dfrac{2}{3}.0,5=\dfrac{1}{3}\left(mol\right)\)

\(\Rightarrow m_{KClO_3}=\dfrac{1}{3}.122,5=40,83\left(g\right)\)

2. \(PTHH:3Fe+2O_2\overset{t^o}{--->}Fe_3O_4\)

Ta có: \(n_{Fe_3O_4}=\dfrac{2,32}{232}=0,01\left(mol\right)\)

a. Theo PT: \(n_{Fe}=3.n_{Fe_3O_4}=0,01.3=0,03\left(mol\right)\)

\(\Rightarrow m_{Fe}=0,03.56=1,68\left(g\right)\)

b. Theo PT: \(n_{O_2}=2.n_{Fe_3O_4}=2.0,01=0,02\left(mol\right)\)

\(\Rightarrow V_{O_2}=0,02.22,4=0,448\left(lít\right)\)

\(n_P=\dfrac{7,44}{31}=0,24mol\)

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

0,24   0,3      0,12

\(V_{O_2}=0,3\cdot22,4=6,72l\)

\(2KClO_3\underrightarrow{t^o}2KCl+3O_2\)

0,2                           0,3

\(m_{KClO_3}=0,2\cdot122,5=24,5g\)

a. \(n_P=\dfrac{7.44}{31}=0,24\left(mol\right)\)

PTHH : 4P + 5O₂ ------t^o----> 2P₂O₅

            0,24    0,3

b. \(V_{H_2}=0,3.22,4=6,72\left(l\right)\)

c. 2KClO₃ ---t^o---> 2KCl + 3O₂

       0,2                              0,3

\(m_{KClO_3}=0,2.122,5=24,5\left(g\right)\)