\(2KClO_3\xrightarrow[]{t^o}2KCl+3O_2\)

\(n_{O_2}=\dfrac{V_{O_2}}{22,4}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

Theo PTHH: \(n_{KClO_3}=\dfrac{0,25.2}{3}\approx0,17\left(mol\right)\)

Vậy muốn điều chế 5,6 lít O₂ cần dùng số gam Kali clorat:

\(m_{KClO_3}=n_{KClO_3}.M_{KClO_3}=0,17.122,5=20,825g\)

             \(n_{O2}\)=\(\dfrac{V}{22,4}\)=\(\dfrac{5,6}{22,4}\)=0,25 (mol)

     PT : 2KClO3 →^to  2KCl + 3O2

số mol:      \(\dfrac{1}{6}\)      ←    \(\dfrac{1}{6}\)      ← 0,25

⇒ m_KClO3 = n . M = \(\dfrac{1}{6}\) . 122,5 ∼_∼ 20,41(g)

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

PTHH: 3Fe + 2O₂ --t^o--> Fe₃O₄

            0,03<-0,02<------0,01

=> m_Fe = 0,03.56 = 1,68 (g)

b) V_O2 = 0,02.22,4 = 0,448 (l)

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

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

\(\Rightarrow\left\{{}\begin{matrix}n_{O_2}=0,02\left(mol\right)\\n_{Fe}=0,03\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{Fe}=0,03\cdot56=1,68\left(g\right)\\V_{O_2}=0,02\cdot22,4=0,448\left(l\right)\end{matrix}\right.\)

\(n_{Fe_3O_4}=\dfrac{m}{M}=\dfrac{4,64}{232}=0,02mol\)

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

0,06   0,04             0,02   ( mol )

\(m_{Fe}=n_{Fe}.M_{Fe}=0,06.56=3,36g\)

\(V_{O_2}=n_{O_2}.22,4=0,04.22,4=0,896l\)

n_Fe3O4 = 2,32/232 = 0,01 mol

3Fe  +  2O₂ ➝ Fe₃O₄

 0,03    0,02      0,01    (mol)

a) m_Fe = 0,03.56 = 1,68 gam

b) V_O2 = 0,02.22,4 = 0,448 lít

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

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

\(\Rightarrow\left\{{}\begin{matrix}n_{Fe}=0,03\left(mol\right)\\n_{O_2}=0,02\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{Fe}=0,03\cdot56=1,68\left(g\right)\\V_{O_2}=0,02\cdot22,4=0,448\left(l\right)\end{matrix}\right.\)

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

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

                0,3<------------------------------0,15

            2KClO₃ --t^o--> 2KCl + 3O₂

               0,1<-----------------0,15

=> \(\left\{{}\begin{matrix}m_{KMnO_4}=0,3.158=47,4\left(g\right)\\m_{KClO_3}=0,1.122,5=12,25\left(g\right)\end{matrix}\right.\)

=> Dùng KClO₃ sẽ cần khối lượng nhỏ hơn

\(n_{O_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

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

Theo PT: \(n_{KClO_3}=\dfrac{2}{3}n_{O_2}=\dfrac{1}{6}\left(mol\right)\Rightarrow m_{KClO_3}=\dfrac{1}{6}.122,5=\dfrac{245}{12}\left(g\right)\)

a. \(n_{O_2}=\dfrac{22.4}{22.4}=1\left(mol\right)\)

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

                 2                                                         1

\(m_{KMnO_4}=2.158=316\left(g\right)\)

b. PTHH : C + O₂ ---t^o--->CO₂ 

                  1     1               1

\(m_{CO_2}=1.44=44\left(g\right)\)

a. n O 2 = 22.4 22.4 = 1 ( m o l ) PTHH : 2KMnO4 ----to----> K2MnO4 + MnO2 + O2 2 1 m K M n O 4 = 2.158 = 316 ( g ) b. PTHH : C + O2 ---to--->CO2 1 1 1 m C O 2 = 1.44 = 44 ( g )