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

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

4K + O₂ --^to--➢ 2K₂O (1)

\(n_{K_2O}=\dfrac{16,8}{94}=\dfrac{42}{235}\left(mol\right)\)

Theo PT1: \(n_{O_2}=\dfrac{1}{2}n_{K_2O}=\dfrac{1}{2}\times\dfrac{42}{235}=\dfrac{21}{235}\left(mol\right)\)

2KClO₃ --^to--➢ 2KCl + 3O₂ (2)

\(n_{O_2\left(1\right)}=n_{O_2\left(2\right)}\)

Theo PT2: \(n_{KClO_3}=\dfrac{2}{3}n_{O_2}=\dfrac{2}{3}\times\dfrac{21}{235}=\dfrac{14}{235}\left(mol\right)\)

\(\Rightarrow m_{KClO_3}=\dfrac{14}{235}\times122,5=7,3\left(g\right)\)

2Mg+O₂-to>2MgO

0,15---0,075 mol

2KClO₃-to->2KCl+3O₂

0,05-----------------------0,075

n _Mg=\(\dfrac{3,6}{24}\)=0,15 mol

=>V_O2=0,075.22,4=1,68l

=>m _KClO3=0,05.122,5=6,125g

Sao lại cho Mg vào mà ''khối lượng kali clorat'' là sao em

 a. \(n_{Mg}=\dfrac{3.6}{24}=0,15\left(mol\right)\)

PTHH : 2Mg + O₂ ----t^o----> 2MgO

             0,15   0,075             

\(V_{O_2}=0,075.22,4=1,68\left(l\right)\)

b. PTHH : 2KClO₃ \(\xrightarrow[MnO_2]{t^o}\) 2KCl + 3O₂

                   0,05                         0,075

\(m_{KClO_3}=0,05.122,5=6,125\left(g\right)\)

\(n_{H_2}=\dfrac{11.2}{22.4}=0.5\left(mol\right)\)

\(a.\)

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

\(0.5.......0.25\)

\(m_{O_2}=0.25\cdot32=8\left(g\right)\)

\(b.\)

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

\(0.5............................................0.25\)

\(m_{KMnO_4}=0.5\cdot158=79\left(g\right)\)

mMg = 3.6/24 = 0.15 (mol) 

2Mg + O2 -to-> 2MgO 

0.15__0.075____0.15 

mMgO= 0.15*40 = 6 (g) 

VO2  = 0.075*22.4 = 1.68 (l) 

2KClO3 -to-> 2KCl + 3O2 

0.05_______________0.075 

mKClO3 = 0.05*122.5 = 6.125 (g) 

PTHH: \(2Mg+O_2\underrightarrow{t^o}2MgO\)

a+b) Ta có: \(n_{Mg}=\dfrac{3,6}{24}=0,15\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{O_2}=0,075\left(mol\right)\\n_{MgO}=0,15\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{O_2}=0,075\cdot22,4=1,68\left(l\right)\\m_{MgO}=0,15\cdot40=6\left(g\right)\end{matrix}\right.\)

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

Theo PTHH: \(n_{KClO_3}=0,05\left(mol\right)\)

\(\Rightarrow m_{KClO_3}=0,05\cdot122,5=6,125\left(g\right)\)

4P (0,2 mol) + 5O₂ (0,25 mol) \(\underrightarrow{t^o}\) 2P₂O₅.

a) Thể tích không khí cần dùng ở đktc là 0,25.22,4.5=28 (lít).

b) 2KMnO₄ (0,5 mol) \(\underrightarrow{t^o}\) K₂MnO₄ + MnO₂ + O₂ (0,25 mol).

Khối lượng cần tìm là 0,5.158=79 (g).

a.\(n_P=\dfrac{6,2}{31}=0,2mol\)

\(4P+5O_2\rightarrow\left(t^o\right)2P_2O_5\)

0,2   0,25                          ( mol )

\(V_{kk}=V_{O_2}.5=\left(0,25.22,4\right).5=28l\)

b.\(2KMnO_4\rightarrow\left(t^o\right)K_2MnO_4+MnO_2+O_2\)

       0,5                                                      0,25  ( mol )

\(m_{KMnO_4}=0,5.158=79g\)

nK=0,2(mol)

PTHH: 4K + O2 -to-> 2 K2O

nK2O= 0,1(mol) => mK2O=0,1.94=9,4(g)

nO2=0,05(mol) -> V(O2,đktc)=0,05.22,4=1,12(l)

V(kk,dktc)=5.V(O2,dktc)=5.1,12=5,6(l)