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4 ý cuối :
1)
Cu + 2H2SO4→ CuSO4+ SO2+2H2O
Cu0 →Cu+2 +2e║ x1
S+6+2e →S+4 ║ x1
2)
2Al+ 4H2SO4→ Al2(SO4)3+ S+ 4H2O
2Al0→2Al+3 +6e║x1
S+6 +6e→S0 ║x1
3)
4Zn +5H2SO4→ 4ZnSO4+ H2S+ 4H2O
Zn0\(\rightarrow\) Zn+2 +2e ║x4
S+6 +8e →S−2 ║x1
4)
8Fe+ 15H2SO4→ 4Fe2(SO4)3+3H2S+ 12H2O
2Fe0→ 2Fe+3+6e║x4
S+6 +8e →S−2 ║x3
6 ý đầu
1.\(\overset{-3}{4NH_2}+\overset{0}{5O_2}\rightarrow\overset{+2+6}{4NO}+\overset{-2}{6H_2O}\)
4 X \(||\) N-3 + 5e → N+2
5 X \(||\) 2O0 + 4e → 2O-2
2.\(\overset{-3}{4NH3}+\overset{0}{3O_2}\rightarrow\overset{0}{2N_2}+\overset{-2}{6H_2O}\)
2 X \(||\) 2N-3 + 6e → 2N0
3 X \(||\) 2O0 + 4e → 2O-2
3.\(\overset{0}{3Mg}+\overset{+5}{8NO_3}\rightarrow\overset{+2}{3Mg\left(NO_3\right)_2}+\overset{+2}{2NO}+\overset{ }{4H_2O}\)
3 X \(||\) Mg0 → Mg+2 + 2e
2 X \(||\) N+5 + 3e → N+2
4.\(\overset{0}{Al}+\overset{+5}{6NO_3}\rightarrow\overset{+3}{Al\left(NO_3\right)_3}+\overset{+4}{3NO_2}+\overset{ }{3H_2O}\)
1 X \(||\) Al0 → Al+3 + 3e
3 X \(||\) N+5 + 1e → N+4
5.\(\overset{0}{Zn}+\overset{+5}{4HNO_3}\rightarrow\overset{+3}{Fe\left(NO_3\right)_3}+\overset{+2}{NO}+\overset{ }{2H_2O}\)
1 X \(||\) Zn0 → Mg+2 + 2e
2 X \(||\) N+5 + 3e → N+4
6.\(\overset{0}{Fe}+\overset{+5}{4HNO_3}\rightarrow\overset{+3}{Fe\left(NO_3\right)_3}+\overset{+2}{NO}+\overset{ }{2H_2O}\)
1 X \(||\) Fe0 → Fe+3 + 3e
1 X \(||\) N+5 + 3e → N+2
a. (1) 2KClO3 \(\underrightarrow{t}\) 3O2 + 2KCl
(2) 5O2 + 4P → 2P2O5
(3) P2O5 + H2O → 2H3PO4
b. (1) BaCO3 → BaO + CO2
(2) BaO + H2O → Ba(OH)2
Nhường e: ( Fe2+ ----> Fe3+ + 1e ) *14
Nhận e: S+6 +2e ----> S+4 (Tỉ lệ 1:2)
2S+6 + 12e ----> 2S0
=> 3S+6 +14e -----> S+4 + 2S0
Cân bằng: 14FeO + 48H+ + 3SO42- →14Fe3+ + SO2 + 2S + 24H2O
a/
\(2HCl\underrightarrow{^{đpdd}}H_2+Cl_2\)
\(3Cl_2+2Fe\underrightarrow{^{to}}2FeCl_3\)
\(FeCl_3+3NaOH\rightarrow3NaCl+Fe\left(OH\right)_3\)
\(2NaCl+H_2SO_4\underrightarrow{^{to}}Na_2SO_4+2HCl\)
\(HCl+CuO\rightarrow CuCl_2+H_2O\)
\(CuCl_2+AgNO_3\rightarrow AgCl+Cu\left(NO_3\right)2\)
b/
\(2HCl\underrightarrow{^{đpdd}}H_2+Cl_2\)
\(Cl_2+2Na\underrightarrow{^{to}}2NaCl\)
\(2NaCl+H_2SO_{4_{dac}}\underrightarrow{^{to}}Na_2SO_4+2HCl\)
\(2HCl+Fe\rightarrow FeCl_2+H_2\)
c/
\(MnO_2+4HCl_đ\underrightarrow{^{to}}MnO_2+Cl_2+2H_2O\)
\(Cl_2+2K\underrightarrow{^{to}}2KCl\)
\(2KCl+H_2SO_4\underrightarrow{^{to}}K_2SO_4+2HCl\)
\(2HCl\underrightarrow{^{đpdd}}H_2+Cl_2\)
\(Cl_2+2NaBr\rightarrow2NaCl+Br_2\)
\(Br_2+2NaI\rightarrow NaBr+I_2\)
d/
\(2KMnO_4+16HCl_đ\underrightarrow{^{to}}2KCl+2MnCl_2+5Cl_2+8H_2O\)
\(Cl_2+H_2\underrightarrow{^{as}}2HCl\)
\(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)
\(FeCl_3+3AgNO_3\rightarrow3AgCl+Fe\left(NO_3\right)_3\)
Mg\(\rightarrow\)Mg+2 +2e
N+5+4e \(\rightarrow\)NO+NO2
2Mg+6HNO3\(\rightarrow\)2Mg(NO3)2+NO+NO2+3H2O
a) 4HCl + MnO2 --> MnCl2 + Cl2 +2H2O
2Fe + 3Cl2 ---> 2FeCl3
FeCl3 + 3NaOH --> Fe(OH)3 + 3NaCl
NaCl + H2SO4 đ---> NaHSO4 + HCl
2HCl + CuO ---> CuCl2 + H2O
CuCl2 +2AgNO3 ---> Cu(NO3)2 + 2AgCl
b) 2KMnO4 + 16HCl ---> 2KCl + 2MnCL2 +5Cl2 +8H2O
Cl2 + H2--->2HCl
6HCl + Fe2O3 ---> 2FeCl3 +3H2O
FeCl3 + 3AgNO3 --> Fe(NO3)3 +3AgCl
2AgCl --to---> 2Ag + Cl2
Cl2 + 2NaBr ---> 2NaCl + Br2
Br2 + 2NaI --> 2NaBr + I2
I2 +Zn --to--> ZnI2
ZnI2 + 2NaOH --> Zn(OH)2 +2NaI
c) 2KCl ---dpnc--> 2K + Cl2
Cl2 + 2KOH --> KCl + KClO + H2O
4KClO --> KClO3 +3 KCl
4KClO3 ---> 3KClO4 + KCl
3KClO4 + 8Al ---> 4Al2O3 + 3KCl
KCl + AgNO3 --> AgCl + KNO3
d) 3Cl2 + 6KOH ---> KClO3 + 5KCl +3H2O
2KClO3 ---> 2KCl +#O2
2KCl --> 2K + Cl2
2Cl2 + Ca(OH)2 ---> CaCl2 +Ca(ClO)2
Ca(ClO)2 ---> CaCl2 + O2
CaCl2 ---> Ca + Cl2
Cl2 ra O2 ????
e)6HCl + KClO3 ---> KCl +3Cl2 +3H2O
3Cl2 +6KOH --> 5KCl + KClO +3 H2O
2KClO3 --> 2KCl + 3O2
2KCl --> 2K + CL2
CL2 +H2 --> 2HCl
2HCl +Fe--> FeCl2 + H2
Cl2 + 2FeCl2 -->2FeCl3
FeCl3 +3NaOH --> Fe(OH)3 +3NaCl
4Mg +5H2SO4 đặc nóng =>4MgSO4 + H2S +4H2O
Tổng hệ số nguyên tối giản của các chất trong pt là 4+5+4+1+4=16
Tâm ơi , bn cộng nhầm rùi , là 18 chứ ko phải 16 .
cảm ơn bn đã trả lời câu hỏi giúp mình nhé .
