Fe\(_x\)O\(_y\) + HCL \(\rightarrow\)FeCl\(_{\dfrac{2y}{x}}\) + H\(_2\)O
Fe\(_x\)O\(_y\) + H\(_2\)SO\(_4\) \(\rightarrow\)Fe(NO\(_3\))\(_3\) + S + H\(_2\)O
trước là SO4 mà sao sau phản ứng lại xuất hiện NO3 hay z
Fe\(_x\)O\(_y\) + H\(_2\)SO\(_4\) \(\rightarrow\)Fe\(_2\)(SO\(_4\))\(_3\) + SO\(_2\) + H\(_2\)O
2FexOy + (6x-2y)H2SO4 -> xFe2(SO4)3 + (3x-2y)SO2 + (6x-2y)H2O
C\(_x\)O\(_y\)(COOH)\(_2\) +O\(_2\) \(\rightarrow\) CO\(_2\) + H\(_2\)O
CxHy(COOH)2 + (x+\(\dfrac{y}{4}\)+\(\dfrac{5}{2}\))O2 -> (x+2)CO2 + (\(\dfrac{y}{2}\)+1)H2O
O\(_2\)\(\underrightarrow{ }\)Fe\(_3\)O\(_4\)\(\underrightarrow{ }\)Fe\(\underrightarrow{ }\)FeCl\(_2\)
\(\downarrow\)
H\(_2\)
(1)O\(_2\)+ \(\rightarrow\)Fe\(_3\)O\(_4\)
2O2+3Fe-to>Fe3O4
Fe3O4+4H2-to>3Fe+4H2O
2Fe+3Cl2-to>2Fecl3
Fe+2HCl->Fecl2+H2
\(3Fe+2O_2\rightarrow\left(t^o\right)Fe_3O_4\)
\(Fe_3O_4+4H_2\rightarrow\left(t^o\right)3Fe+4H_2O\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
Cân bằng phương trình sau:
a, Zn + O\(_2\)-->ZnO
b, SO\(_2\) + O\(_2\)-->SO\(_3\)
c, Ca+H\(_2\)S0\(_4\)--->CaSO\(_4\)+H\(_2\)O
d, HCl+MnO\(_2\)--->MnCl\(_2\)+Cl\(_2\)+H\(_2\)O
e, Ba(OH)\(_2\)+HCl--->BaCl\(_2\)+H\(_2\)O
f, Fe+H\(_2\)PO\(_{4đ.n}\)--->Fe\(_2\)(SO\(_4\))\(_3\)+SO\(_2\)+H\(_2\)O
g, Fe\(_x\)O\(_y\)+CO--->Fe+CO\(_2\)
h, K\(_2\)O+H\(_3\)PO\(_4\)--->K\(_3\)PO\(_4\)+H\(_2\)O
i, C\(_x\)H\(_y\)+O\(_2\)--->CO\(_2\)+H\(_2\)O
k, FeS\(_2\)+O\(_2\)---> Fe\(_2\)O\(_3\)+SO\(_2\)
a> 2Zn+O2-->2ZnO ( tỉ lệ hệ số là 2:1:2)
b>2:1:2
d>4:1:1:1:2
e>1:2:1:2
f>2:6:1:3:6
k> 2:11/2:1:4
h> 3K2O+ 2H3PO4-->2K3PO4+ 3H2O
g> FexOy + yCO--->xFe+ yCO2
i>CxHy+ (x+y/2)o2--> x CO2+yH2O
cân bằng PTHH:
d) Fe x O y + HNO 3 → Fe(NO 3 ) 3 + NO + H 2 O
e) Fe x O y + HNO 3 → Fe(NO 3 ) 3 + NO 2 + H 2
f) Fe x O y + HCl → FeCl 2y/x + H 2 O
g) Fe x O y + H 2 SO 4 → Fe 2 (SO 4 ) 2y/x + H 2 O
1,Hoàn thành phương trình hóa học sau:
a,FexOy + CO\(\rightarrow\) Fe +CO2.
b,FexOy + HCl \(\rightarrow\) FeCl2y/x + H2O.
c,Fe + HNO3 \(\rightarrow\) Fe ( NO3)3 + NO + H2O.
d,Al + H2SO4 \(\rightarrow\) Al2 ( SO4)3 + SO2 + H2O.
a,X Là 2 ;Y Là 3
b,X Là 2 ;Y Là 2
c,3Fe+2HNO3 tạo ra Fe(NO3)3+2NO3+H2O
d,2Al+H2SO4 tạo ra Al2(SO4)3 +SO2+H2O
a) FexOy + yCO → xFe + yCO2
b) FexOy + 2yHCl → xFeCl\(\dfrac{2y}{x}\) + yH2O
c) Fe + 4HNO3 → Fe(NO3)3 + NO + 2H2O
d) 2Al + 6H2SO4 → Al2(SO4)3 + 3SO2 + 6H2O
a)FexOy + (y-x)CO ---> xFeO + (y-x)CO2
b)FexOy + 2y HCL --> x FeCl\(\dfrac{2y}{x}\) + y H2O
c) Fe + 4HNO3 → Fe ( NO3)3 + NO + 2H2O.
d)4Al + 9H2SO4 → 2Al2 ( SO4)3 + 3SO2 + 9H2O.
hãy thực hiện chuyển đổi hóa học sau:
a) K→K\(_2\)O→KOH
b) P→P\(_2\)O\(_3\)→H\(_3\)
c) Fe→Fe\(_3\)O\(_4\)→Fe→FeCL\(_2\)
\(a,4K+O_2\rightarrow2K_2O\\ K_2O+H_2O\rightarrow2KOH\\ b,4P+3O_{2\left(thiếu\right)}\rightarrow2P_2O_3\\ P_2O_3+3H_2O\rightarrow2H_3PO_3\\ c,3Fe+2O_2\rightarrow\left(t^o\right)3Fe_3O_4\\ Fe_3O_4+8Al\rightarrow\left(t^o\right)9Fe+4Al_2O_3\\ Fe+2HCl\rightarrow FeCl_2+H_2\)
Cân bằng số nguyên tử mỗi nguyên tố :
a/ CxHy + O2 - - - -> CO2 + H2O
b/ FexOy + HCl - - - -> FeCl2y /x + H2O
c/ Fe3O4 + HCl - - -> FeCl2 + FeCl3 +H2O
a) CxHy + O2 ------->CO2 +H2O
Sau khi cân bằng là:
CH4 + 2O2 -> CO2 + 2H2O
b)Sau khi cân bằng là:
Fe2O3 + 6HCl -> 2FeCl3 + 3H2O
c)Sau khi cân bằng là:
Fe3O4 + 8HCl -> 2FeCl3 + FeCl2 +4H2O
a/ CH2 + 3O2 ----> CO2 + 2H2O
b/ 2FeO + 4HCl -----> 2FeCl2 + 2H2O
c/ Fe3O4 + 8HCl ------> FeCl2 + 2FeCl3 + 4H2O