\(1.P_2O_5+3H_2O\rightarrow2H_3PO_4\)

\(2.K_2O+H_2O\rightarrow2KOH\)

\(3.Fe_3O_4+8HCl\rightarrow FeCl_2+2FeCl_3+4H_2O\)

\(4.Na_2O+SO_2\rightarrow Na_2SO_3\)

\(5.P_2O_5+6NaOH\rightarrow2Na_3PO_4+3H_2O\)

\(6.Ba\left(OH\right)_2+H_2SO_4\rightarrow BaSO_4+2H_2O\)

\(7.Fe\left(OH\right)_2+H_2SO_4\rightarrow FeSO_4+2H_2O\)

\(8.Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\)

bài 1

a,

(1) 2Al + 6HCl-----> 2AlCl3 + 3H2

(2)AlCl3 + 3NaOH ----> Al(OH)3 + 3NaCl

(3)2Al(OH)3 + 3H2SO4 -----> Al2(SO4)3 + 6H2O

(4)Al2(SO4)3 + 3BaCl2 -------> 2AlCl3 + 3BaSO4

(5) 4AlCl3 + 3O2 --------> 2Al2O3 + 6Cl2

(6)2Al2O3 -------đpnóng chảy, criotit------->4 Al + 3O2

b,

(1) 2Fe + 3Cl2 --------------> 2FeCl3

(2)FeCl3 + 3NaOH -------------> Fe(OH)3 + 3NaCl

(3)2Fe(OH)3 + 3H2SO4 --------------> Fe2(SO4)3 + 6H2O

(4)Fe2(SO4)3 + 3Pb(NO3)2--------------> 2Fe(NO3)3 + 3PbSO4

(5)4Fe(NO3)3 -------------> 2Fe2O3 + 12NO2 + 3O2

(6) Fe2O3 + 3H2 ------> 2Fe + 3H2O

bài dài và nhàm quá

1)

\(2Fe\left(OH\right)_3\) -> \(Fe_2O_3\) + \(3H_2O\)

\(Fe_2O_3\) + \(3CO\) -> \(2Fe\) + \(3CO_2\)

\(2Fe+3Cl_2\) -> \(2FeCl_3\)

\(FeCl_3+3AgNO_3\) -> \(3AgCl\) +\(Fe\left(NO_3\right)_3\)

2)

\(4FeS+7O_2\) -> \(2Fe_2O_3+4SO_2\)

\(Fe_2O_3+3CO\) -> \(2Fe+3CO_2\)

\(2Fe+3Cl\) ->\(2FeCl_3\)

\(2FeCl_3+3Ba\left(OH\right)_2\)->\(3BaCl_2+2Fe\left(OH\right)_3\)

3)

\(2Fe+3Cl_2\) -> \(2FeCl_3\)

\(2FeCl_3+3Ba\left(OH\right)_2\) -> \(3BaCl_2+2Fe\left(OH\right)_3\)

\(2Fe\left(OH\right)_3\) -> \(Fe_2O_3+3H_2O\)

\(Fe_2O_3+3CO\) -> \(2Fe+3CO_2\)

4)

\(2Cu+O_2\) -> \(2CuO\)

\(CuO+2HCl\) -> \(CuCl_2+2H_2O\)

\(CuCl_2+2NaOH\) -> \(2NaCl_2+Cu\left(OH\right)_2\)

\(Cu\left(OH\right)_2+H_2SO_4\) -> \(2H_2O+CuSO_4\)

24.

xFe^+2y/x --------------> xFe⁺³ + (3x - 2y)e (x3)

N^{+ 5} + 3e -------------> N⁺² (x(3x-2y))

=> 3xFe^+2y/x + (3x-2y)N⁺⁵ -----> 3xFe⁺³ + (3x-2y)N⁺²

Vậy pư cân bằng như sau:

3Fe_xO_y + (12x - 2y)HNO₃ ----> 3xFe(NO₃)₃ + (3x-2y)NO + (6x-y)H₂O

25.

xFe^+2y/x --------------> xFe⁺³ + (3x - 2y)e -----------------(x1)

N^{+ 5} + 1e -------------> N⁺⁴ ----------------------- (x(3x-2y))

=> xFe^+2y/x + (3x-2y)N⁺⁵ -----> xFe⁺³ + (3x-2y)N⁺²

Vậy pư cân bằng như sau:

Fe_xO_y + (6x - 2y)HNO₃ ----> xFe(NO₃)₃ + (3x-2y)NO₂ + (3x-y)H₂O

a)(5x-2y)Fe+(18x-6y)HNO₃--->(5x-2y)Fe(NO₃)₃+3N_xO_y+(9x-3y)H₂O

b)(5x-2y)Mg+(12x-4)HNO₃--->(5x-2y)Mg(NO₃)₂+2N_xO_y+(6x-2y)H₂O

c)2Fe_xO_y+(6x-2y)H₂SO₄--->xFe₂(SO₄)₃+(3x-2y)SO₂+(6x-2y)H₂O

d)Fe_xOy+(6x-2y)HNO₃--->xFe(NO₃)₃+(3x-2y)NO₂+(3x-y)H₂O

e)2Crl₃+27Cl₂+64KOH--->2K₂CrO₄+6KlO₄+54KCl+32H₂O

f)3Cu₂S+16HNO₃--->10NO+3Cu(NO₃)₂+3CuSO₄+8H₂O

Kiểm tra lại nhé!!!

Bạn tự cân bằng các phương trình nhé

Fe+Cl2->FeCl3(đk nhiệt độ)

FeCl3+NaOH->NaCl+Fe(OH)3

Fe(OH)3+H2SO4->Fe2(SO4)3+H2O

Fe(OH)3->Fe2O3+H2O(điều kiện nhiệt độ)

Fe2O3+CO(dư)->Fe+CO2(đk nhiệt độ)

b)Al+Cl2->AlCl3

AlCl3+NaOH(vừa đủ)->Al(OH)3+NaCl

Al(OH)3+H2SO4->Al2(SO4)3+H2O

Al(OH)3->Al2O3+H2O(đk nhiệt độ)

Al2O3->Al+O2(điện phân nóng chảy criolit)

S+O2->SO2(đk nhiệt độ)

SO2+Br2+H2O->H2SO4+HBr

H2SO4+BaO->BaSO4+H2O

H2SO4(đặc)+Cu->CuSO4+SO2+H2O

1) Cu + 4HNO₃ → Cu(NO₃)₂ + 2NO₂ + 2H₂O

2) 2FeO + 4H₂SO₄ → Fe₂(SO₄)₃ + SO₂ + 4H₂O

3) Fe₃O₄ + 10HNO₃ → 3Fe(NO₃)₃ + NO₂ + 5H₂O

4) 8Fe + 27HNO₃ → 8Fe(NO₃)₃ + 3NH₃ + 9H₂O

Cảm ơn 😊😊😊

a, nFexOy+ \(\dfrac{xm-yn}{2}\)O2---> xFenOm

b, Al2O3 + 6NaHSO4 ----> Al2(SO4)3 + 3Na2SO4 + 6H2O

c, 2Fe3O4+ 10H2SO4 đặc-----> 3Fe2(SO4)3 + SO2 + 10H2O

d, Ca3(PO4)2 + 4H3PO4 -----> 3Ca(H2PO4)2

a)

\(2Ca+O_2\rightarrow2CaO\)

\(CaO+H_2O\rightarrow Ca\left(OH\right)_2\)

\(Ca\left(OH\right)_2+CO_2\rightarrow CaCO_3+H_2O\)

\(CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2\)

\(CaCl_2+2AgNO_3\rightarrow Ca\left(NO_3\right)_2+2AgCl\downarrow\)

b)

\(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)

\(FeCl_3+3KOH\rightarrow3KCl+Fe\left(OH\right)_3\downarrow\)

\(2Fe\left(OH\right)_3\underrightarrow{t^o}Fe_2O_3+3H_2O\)

\(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\)

)2Ca + O₂ ------>2CaO

CaO + H₂O --->Ca(OH)₂

Ca(OH)2+CO₂ -->CaCO₃+H₂OO

CaCO₃+2HCl--->CaCl₂+H₂CO₃

CaCl₂+2HNO₃--->Ca(NO3)₂+2HCl

B,tự giải nha