a.\(n_{KClO_3}=\dfrac{m_{KClO_3}}{M_{KClO_3}}=\dfrac{12,25}{122,5}=0,1mol\)

\(2KClO_3\rightarrow\left(t^o\right)2KCl+3O_2\)

    2                       2           3       ( mol )

  0,1                                 0,15

\(V_{O_2}=n_{O_2}.22,4=0,15.22,4=3,36l\)

b.\(V_{kk}=V_{O_2}.5=3,36.5=16,8l\)

c.\(n_{Fe}=\dfrac{m_{Fe}}{M_{Fe}}=\dfrac{28}{56}=0,5mol\)

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

  3        2                   1         ( mol )

0,5   >   0,15                          ( mol )

0,225              0,15                 ( mol )

\(m_{Fe\left(du\right)}=n_{Fe\left(du\right)}.M_{Fe}=\left(0,5-0,225\right).56=15,4g\)

Cập nhật lại câu hỏi đi bạn :v

đây là hóa học ak bạn

 a) n_O2 = 4,48 / 22,4 = 0,2 (mol)

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

m _KClO3 = 122,5 * ( 0,2 * 2 / 3 ) =49/3 g

b) mol KCl thu đc

 n_KCl = 1, 5*2 /2 = 1,5 mol

m_KCl = 1,5 *74,5 =111, 75 g

PT: 2KClO₃ ------t*----- > 2KCl + 3O₂

a, Số mol O₂ cần điều chế là: \(n_{O_2}=\frac{4,48}{22,4}=0,2\left(mol\right)\)

Theo PT,ta có: \(n_{KClO_3}=\frac{2}{3}.n_{O_2}=\frac{2}{3}.0,2=\frac{2}{15}\left(mol\right)\)

Khối lượng KClO₃ cần dùng là: \(m_{KClO_3}=\frac{2}{15}.122,5=\frac{49}{3}\approx16,33\left(g\right)\)

b,Theo PT + đb ,ta có: \(n_{KCl}=n_{KClO_3}=1,5\left(mol\right)\)

=> Khối lượng KCl thu đc là: \(m_{KCl}=1,5.74,5=111,75\left(g\right)\)

=.= hok tốt!!

Bài 4 :

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

0,4__________________________0,2

\(n_{KMnO4}=\frac{63,2}{158}=0,4\left(mol\right)\)

\(\Rightarrow V_{O2}=0,2.22,4=4,48\left(l\right)\)

Bài 5:

\(n_{KClO3}=\frac{24,5}{122,5}=0,2\left(mol\right)\)

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

0,2_______________0,3__(mol)

\(n_{KMnO4}=\frac{24,5}{158}=0,155\left(mol\right)\)

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

0,155_________________________0,078__(mol)

\(\Rightarrow V_{O2}=0,078.22,4=1,75\left(l\right)\)

Vậy khi dùng lượng KMnO4 và KClO3 như nhau thì KClO3 điều chế được nhiều O2 hơn

\(n_{KClO_3\left(bd\right)}=\dfrac{55,125}{122,5}=0,45\left(mol\right)\)

=> \(n_{KClO_3\left(pư\right)}=\dfrac{0,45.85}{100}=0,3825\left(mol\right)\)

PTHH: 2KClO₃ --t^o,MnO₂--> 2KCl + 3O₂

             0,3825------------------->0,57375

=> \(V_{O_2}=0,57375.22,4=12,852\left(l\right)\)

2KClO₃-to>2KCl+3O₂

0,45---------------------0,675 mol

n _KClO3=\(\dfrac{55,125}{122,5}\)=0,45 mol

=>H=85%

=>V_O2=0,675.22,4.\(\dfrac{85}{100}\)=12,852l

a. \(n_{Fe_3O_4}=\dfrac{6,96}{232}=0,03\left(mol\right)\)

PTHH : 3Fe + 2O₂ -t^o-> Fe₃O₄

              0,09    0,06        0,03

\(m_{Fe}=0,09.56=5,04\left(g\right)\)

\(V_{O_2}=0,06.22,4=1,344\left(l\right)\)

b. PTHH : 2KCl + 3O₂ -> 2KClO₃

                            0,06           0,04

\(m_{KClO_3}=0,04.122,5=4,9\left(g\right)\)

4

n _Fe3O4=\(\dfrac{6,96}{232}=0,03mol\)

3Fe+2O₂-to>Fe₃O₄

0,09---0,06-----0,03 mol

=>m _Fe=0,09.56=5,04g

=>V_O2=0,06.22,4=1,344l

b) 

2KClO₃-to>2KCl+3O₂

0,04----------------------0,06 mol

=>m _KClO3=0,04.122,5=4,9g

\(n_{Fe_3O_4}=\dfrac{6,96}{232}=0,03\left(mol\right)\\ 3Fe+2O_2\rightarrow\left(t^o\right)Fe_3O_4\\ n_{Fe}=3.0,03=0,09\left(mol\right);n_{O_2}=0,03.2=0,06\left(mol\right)\\ a,\Rightarrow m_{Fe}=0,09.56=5,04\left(g\right);V_{O_2\left(đktc\right)}=0,06.22,4=1,344\left(l\right)\\ 2KClO_3\rightarrow\left(t^o\right)2KCl+3O_2\uparrow\\ n_{KClO_3}=\dfrac{2}{3}.n_{O_2}=\dfrac{2.0,06}{3}=0,04\left(mol\right)\\ \Rightarrow b,m_{KClO_3}=122,5.0,04=4,9\left(g\right)\)

\(n_{O_2}=\dfrac{43.2}{32}=1.35\left(mol\right)\)

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

\(0.9...........................1.35\)

\(H\%=\dfrac{0.9}{1}\cdot100\%=90\%\)

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

______1_____________1,5 (mol)

⇒ m_{O2 (lí thuyết)} = 1,5.32 = 48 (g)

Mà: m_{O2 (thực tế)} = 43,2 (g)

\(\Rightarrow H\%=\dfrac{43,2}{48}.100\%=90\%\)

Bạn tham khảo nhé!

\(n_{O_2}\)=\(\dfrac{43,2}{32}=1,35\left(mol\right)\)

PTHH 2KClO₃-----t^o--->2KCl +3O₂

=>\(n_{O_2\left(lt\right)}=1.\dfrac{3}{2}=1,5\left(mol\right)\)

=>H%=\(\dfrac{1,35}{1,5}.100\%=90\%\)

\(a.2KClO_3-^{t^o}\rightarrow2KCl+3O_2\\ b.n_{O_2}=\dfrac{3}{2}n_{KClO_3}=\dfrac{3}{2}.\dfrac{12,5}{122,5}=\dfrac{15}{98}\left(mol\right)\\ \Rightarrow V_{O_2}=\dfrac{15}{98}.22,4=\dfrac{24}{7}\left(l\right)\approx3,24\left(l\right)\)

\(n_{O2}=\frac{4,032}{22,4}=0,18\left(mol\right)\)

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

________2_________2_______ 3_

________________________0,18__

\(\Rightarrow n_{KClO3}=0,18.\frac{2}{3}=0,12\left(mol\right)\)

\(\Rightarrow m_{KClO3}=0,36.122,5=14,7\left(g\right)\)

\(n_{O2}=\frac{V}{22,4}=\frac{4,032}{22,4}0,18\left(mol\right)\)

PTPU :

2KClO3 -> 3O2 + 2KCl

PU 0,12 <- 0,18 (mol)

Vậy \(m_{KClO3}=n.M=0,12.122,5=14,7\left(g\right)\)

\(a.2KClO_3-^{t^o}\rightarrow2KCl+3O_2\\ n_{O_2}=\dfrac{3}{2}n_{KClO_3}=0,6\left(mol\right)\\ \Rightarrow V_{O_2}=0,6.22,4=13,44\left(l\right)\\ n_{KCl}=n_{KClO_3}=0,4\left(mol\right)\\ \Rightarrow m_{KCl}=0,4.74,5=29,8\left(g\right)\)