Theo gt ta có: $n_{H_2}=0,75(mol)$

a, $2H_2+O_2\rightarrow 2H_2O$

Ta có: $n_{O_2}=0,5.n_{H_2}=0,375(mol)\Rightarrow V_{O_2}=8,4(l)\Rightarrow V_{kk}=42(l)$

b, $2KMnO_4\rightarrow K_2MnO_4+MnO_2+O_2$

Ta có: $n_{KMnO_4}=2.n_{O_2}=0,75(mol)\Rightarrow m_{KMnO_4}=118,5(g)$

a)

\(2H_2 + O_2 \xrightarrow{t^o} 2H_2O\\ V_{O_2} = \dfrac{V_{H_2}}{2} = \dfrac{16,8}{2} = 8,4(lít)\\ V_{không\ khí} = \dfrac{8,4}{20\%} = 42(lít)\)

b)

\(n_{O_2} = \dfrac{8,4}{22,4} = 0,375(mol)\\ 2KMnO_4 \xrightarrow{t^o} K_2MnO_4 + MnO_2 + O_2\\ n_{KMnO_4} = 2n_{O_2} = 0,75(mol)\\ \Rightarrow m_{KMnO_4} = 0,75.158 = 118,5(gam)\\ 2KClO_3 \xrightarrow{t^o} 2KCl + 3O_2\\ n_{KClO_3} = \dfrac{2}{3}n_{O_2} = 0,25(mol)\\ \Rightarrow m_{KClO_3} = 0,25.122,5 = 30,625(gam)\)

\(n_{H_2}=\dfrac{16.8}{22.4}=0.75\left(mol\right)\)

\(2H_2+O_2\underrightarrow{t^0}2H_2O\)

\(0.75...0.375\)

\(V_{O_2}=0.375\cdot22.4=8.4\left(l\right)\)

\(V_{kk}=5V_{O_2}=8.4\cdot5=42\left(l\right)\)

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

\(0.75..............................................0.375\)

\(m_{KMnO_4}=0.75\cdot158=118.5\left(g\right)\)

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

\(0.25.......................0.375\)

\(m_{KClO_3}=0.25\cdot122.5=30.625\left(g\right)\)

a, \(2Zn+O_2\underrightarrow{t^o}2ZnO\)

b, \(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)

Theo PT: \(n_{O_2}=\dfrac{1}{2}n_{Zn}=0,1\left(mol\right)\Rightarrow V_{O_2}=0,1.22,4=2,24\left(l\right)\)

\(\Rightarrow V_{kk}=5V_{O_2}=11,2\left(l\right)\)

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

Theo PT: \(n_{KMnO_4}=2n_{O_2}=0,2\left(mol\right)\Rightarrow m_{KMnO_4}=0,2.158=31,6\left(g\right)\)

a) $n_{Al} = \dfrac{5,4}{27} = 0,2(mol)$
$4Al + 3O_2 \xrightarrow{t^o} 2Al_2O_3$

Theo PTHH : $n_{O_2} = \dfrac{3}{4}n_{Al} = 0,15(mol)$
$V_{O_2} = 0,15.22,4 = 3,36(lít)$

b) $2 KClO_3 \xrightarrow{t^o} 2KCl + 3O_2$

$n_{KClO_3} = \dfrac{2}{3}n_{O_2} = 0,1(mol)$
$m_{KClO_3} = 0,1.122,5 = 12,25(gam)$

\(n_{Al}=\dfrac{m}{M}=\dfrac{5,4}{27}=0,2\left(mol\right)\\ PTHH:4Al+3O_2-^{t^o}>2Al_2O_3\)

tỉ lệ:         4      :     3     :     2

n(mol)     0,2---->0,15---->0,1

\(V_{O_2\left(dktc\right)}=n\cdot22,4=0,15\cdot22,4=3,36\left(l\right)\\ PTHH:2KClO_3-^{t^o}>2KCl+3O_2\)

tỉ lệ:            2             :      2         :    3

n(mol)        0,1<-------------------------0,15

\(m_{KClO_3}=n\cdot M=0,1\cdot\left(39+35,5+16\cdot3\right)=12,25\left(g\right)\)

4Al+3O₂-to>2Al₂O₃

0,2---0,15------0,1 mol

n Al=0,2 mol

VO₂=0,15.22,4=3,36l

b) 2KClO₃-to>2KCl +3O₂

    0,1---------------------0,15 mol

->m KClO₃=0,1.122,5g

#yT

nP = 3,1 : 31 = 0,1 (mol) 
pthh : 4P + 5O2 -t--> 2P2O5 (1)  
           0,1--> 0,125   (mol) 
=> VO2 = 0,125 .22,4 = 2,8(l) 
pthh : 2KMnO4 -t--> K2MnO4 + MnO2 +O2 (2)
                 0,25<---------------------------  0,125(mol) 
=> mKMnO4 = 0,25 .158 = 39,5(g) 
 d ) (1)  là Phản ứng hóa hợp 
        (2) là phản ứng phân hủy

nP = 3,1/31 = 0,1 (mol)

PTHH: 4P + 5O2 -> (t°) 2P2O5 (phản ứng hóa hợp)

Mol: 0,1 ---> 0,125

VO2 = 0,125 . 22,4 = 2,8 (l)

PTHH: 2KMnO4 -> (t°) K2MnO4 + MnO2 + O2 (phản ứng phân hủy)

nKMnO4 = 0,125 . 2 = 0,25 (mol)

mKMnO4 = 0,25 . 158 = 39,5 (g)

a) PTHH: 4P (0,1 mol) + 5O₂ (0,125 mol) \(\underrightarrow{t^o}\) 2P₂O₅.

b) Ở đktc: \(V_{O_2pư}\)=0,125.22,4=2,8 (lít).

c) Điều chế oxi tử KMnO₄:

2KMnO₄ (0,25 mol) \(\underrightarrow{t^o}\) K₂MnO₄ + MnO₂↓ + O₂↑ (0,125 mol).

Khối lượng KMnO₄ cần dùng:

\(m_{KMnO_4}\)=0,25.158=39,5 (g).

d) Hai phản ứng trên đều là phản ứng oxy hóa - khử.

\(n_{H_2}=\dfrac{V}{24,79}=\dfrac{11,2}{24,79}\approx0,45\left(mol\right)\) 

a) \(PTHH:2H_2+O_2\underrightarrow{t^o}2H_2O\) 

                      2        1         2

                    0,45  0,225   0,45

b) \(m_{O_2}=n.M=0,225.\left(16.2\right)=7,2\left(g\right)\\ V_{O_2}=n.24,79=0,225.24,79=5,57775\left(l\right)\) 

c) \(PTHH:2KMnO_4\rightarrow K_2MnO_4+MnO_2+O_2\) 

                      2                       1                1          1

                     0,45               0,225        0,225     0,225

\(m_{KMnO_4}=n.M=0,45.\left(39+55+16.4\right)=71,1\left(g\right).\)

a, \(2H_2+O_2\underrightarrow{t^o}2H_2O\)

b, Ta có: \(n_{H_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)

Theo PT: \(n_{O_2}=\dfrac{1}{2}n_{H_2}=0,25\left(mol\right)\)

\(\Rightarrow m_{O_2}=0,25.32=8\left(g\right)\)

\(V_{O_2}=0,25.22,4=5,6\left(l\right)\)

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

Theo PT: \(n_{KMnO_4}=2n_{O_2}=0,5\left(mol\right)\Rightarrow m_{KMnO_4}=0,5.158=79\left(g\right)\)

a.\(n_{Al_2O_3}=\dfrac{30,6}{102}=0,3mol\)

\(4Al+3O_2\rightarrow\left(t^o\right)2Al_2O_3\)

 0,6     0,45           0,3                ( mol )

\(m_{Al}=0,6.27=16,2g\)

\(V_{O_2}=0,45.22,4=10,08l\)

\(V_{kk}=10,08.5=50,4l\)

b.\(2KClO_3\rightarrow\left(t^o,MnO_2\right)2KCl+3O_2\)

       0,3                                           0,45 ( mol )

\(m_{KClO_3}=0,3.122,5=36,75g\)

c.\(2KClO_3\rightarrow\left(t^o,MnO_2\right)2KCl+3O_2\)

       0,3                                           0,45  ( mol )

\(n_{KClO_3}=\dfrac{0,3}{75\%}=0,4mol\)

\(m_{KClO_3}=0,4.122,5=49g\)

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

b, Ta có: \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)

Theo PT: \(n_{Fe_3O_4}=\dfrac{1}{3}n_{Fe}=\dfrac{1}{30}\left(mol\right)\Rightarrow m_{Fe_3O_4}=\dfrac{1}{30}.232\approx7,733\left(g\right)\)

c, Theo PT: \(n_{O_2}=\dfrac{2}{3}n_{Fe}=\dfrac{1}{15}\left(mol\right)\)

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

\(n_{KMnO_4\left(LT\right)}=2n_{O_2}=\dfrac{2}{15}\left(mol\right)\)

\(\Rightarrow m_{KMnO_4\left(LT\right)}=\dfrac{2}{15}.158=\dfrac{316}{15}\left(g\right)\)

Mà: H% = 85%

\(\Rightarrow m_{KMnO_4\left(TT\right)}=\dfrac{\dfrac{316}{15}}{85\%}\approx24,78\left(g\right)\)

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

PTHH: \(3Fe+2O_2\xrightarrow[]{t^o}Fe_3O_4\)

            0,3--->0,2----->0,1

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

b) \(V_{O_2}=0,2.22,4=4,48\left(l\right)\Rightarrow V_{kk}=4,48.5=22,4\left(l\right)\)

c) \(n_{O_2\left(hao,h\text{ụt}\right)}=0,2.10\%=0,02\left(mol\right)\)

\(\Rightarrow n_{O_2\left(t\text{ổng}\right)}=0,2+0,02=0,22\left(mol\right)\)

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

              0,44<------------------------------------0,22

\(\Rightarrow m_{KMnO_4}=0,44.158=69,52\left(g\right)\)

a)

PTHH: 3Fe + 2O₂  ^{____\(t^o\)____}> Fe₃O₄  (1) 

b) Ta có: n_Fe  = \(\dfrac{25.2}{56}=0.45\left(mol\right)\)

Theo (1): n_{\(O_2\)}= \(\dfrac{2}{3}n_{Fe}=\dfrac{2}{3}0.45=0.3\left(mol\right)\)

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

c) PTHH: 2KClO₃ ^{__\(t^o\)___}>  2KCl + 3O₂   (2)

-Muốn điều chế được lượng oxi dùng cho phản ứng trên thì \(n_{O_2\left(2\right)}=n_{O_2\left(1\right)}=0.3\left(mol\right)\)

Theo (2) \(n_{KClO_3}=\dfrac{2}{3}n_{O_2}=\dfrac{2}{3}0.3=0.2\left(mol\right)\)

=> \(m_{KClO_3}=0.2\cdot122.5=24.5\left(g\right)\)

\(4P + 5O_2 \xrightarrow{t^o} 2P_2O_5\\ n_{O_2} = \dfrac{5}{4}n_P = \dfrac{5}{4}.\dfrac{3,72}{31} = 0,15(mol)\\ 2KClO_3 \xrightarrow{t^o} 2KCl + 3O_2\\ n_{KClO_3} = \dfrac{2}{3}n_{O_2} = 0,1(mol)\\ \Rightarrow m_{KClO_3} = 0,1.122,5 = 12,25(gam)\)