1 a) \(2HgO\rightarrow2Hg+O_2\)

b) \(2Fe\left(OH\right)_3\rightarrow Fe_2O_3+3H_2O\)

c) \(Na_2CO_3+CaCl_2\rightarrow CaCO_3+2NaCl\)

2a) \(2Cu+O_2\rightarrow2CuO\)

b) \(N_2+3H_2\rightarrow2NH_3\)

c) \(Fe+2HCl\rightarrow FeCl_2+H_2\)

d) \(Mg\left(OH\right)_2\rightarrow MgO+H_2O\)

C. CaCl₂ , Al(OH)₃ , HNO₃ , Ba₃(PO₄)₂

1. \(FeO+2HCl\rightarrow FeCl_2+H_2O\)

2. \(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)

3. \(2Al+6H_2SO_4\left(đ\right)\underrightarrow{t^o}Al_2\left(SO_4\right)_3+3SO_2+6H_2O\)

4. \(CaCO_3\rightarrow CaO+CO_2\)

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

6. \(3Mg+8HNO_3\rightarrow3Mg\left(NO_3\right)_2+2NO+4H_2O\)

7. \(Na_2CO_3+Ca\left(OH\right)_2\rightarrow CaCO_3+2NaOH\)

8. \(Fe_2O_3+3CO\rightarrow2Fe+3CO_2\)

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

10. \(Fe_3O_4+4H_2SO_4\rightarrow FeSO_4+Fe_2\left(SO_4\right)_3+4H_2O\)

1. FeO + 2HCl ----> FeCl₂ + H₂O

2. CuO + H₂SO₄ ----> CuSO₄ + H₂O

3. 2Al + 6H₂SO₄ --t⁰--> Al₂(SO₄)₃ + 6H₂O + 3SO₂

4. CaCO₃----> CaO + CO₂

5. CaCO₃ + 2HCl----> CaCl₂ + CO₂ + H₂O

6. 3Mg + 8HNO₃ ----> 3Mg(NO₃)₂ + 2NO + 4H₂O

7. Na₂CO₃ + Ca(OH)₂ ----> CaCO₃ + 2NaOH

8. Fe₂O₃ + 3CO ----> 2Fe + 3CO₂

9. Fe₂O₃ + 6HCl ----> 2FeCl₃ + 3H₂O

10. Fe₃O₄ + 4H₂SO₄ ----> FeSO₄ + Fe₂(SO₄)₃ + 4H₂O

Điền các hệ số thích hợp để hoàn thành các PTHH hóa học sau:

1. 4H2 + O2 ---> 2H2O

2. 2Al + 3O2 ---> 2Al2O3

3. Fe + 2HCl ---> FeCl2 + H2

4. Fe2O3 + 3H2 ---> 2Fe + 3H2O

5. 2NaOH + CuSO4 ---> Na2SO4 + Cu(OH)2

6. H2SO4 + 2KOH ---> K2SO4 + 2H2O

7. 3AgNO3 + FeCl3 ---> 3AgCl + Fe(NO3)3

8. CaCO3 + 2HCl ---> CaCl2 + H2O + CO2

9. 4P + 5O2 ---> 2P2O5

10. 2Fe + 3Cl2 ---> 2FeCl3

Điền các hệ số thích hợp để hoàn thành các PTHH hóa học sau:

1. 2H₂ + O₂ ---> 2H₂O

2. 4Al + 3O₂ ---> 2Al₂O3

3. Fe + 2HCl ---> FeCl₂ + H₂

4. Fe₂O₃ + 3H₂ ---> 2Fe + 3H₂O

5. 2NaOH + CuSO₄ ---> Na₂SO₄ + Cu(OH)₂

6. H₂SO₄ + 2KOH ---> 2K₂SO₄ + H₂O

7. 3AgNO₃ + FeCl₃ ---> 3AgCl + Fe(NO₃)₃

8. CaCO₃ + 2HCl ---> CaCl₂ + H₂O + CO₂

9. 4P + 4O₂ ---> 2P₂O₅

10. 2Fe + 3Cl₂ ---> 2FeCl₃

1. 2H2 + O2 ---> 2H2O

2. 4Al + 3O2 ---> 2Al2O3

3. Fe + 2HCl ---> FeCl2 + H2

4. Fe2O3 + 3H2 ---> 2Fe + 3H2O

5. 2NaOH + CuSO4 ---> Na2SO4 + Cu(OH)2

6. H2SO4 + 2KOH ---> 2K2SO4 + H2O

7. 3AgNO3 + FeCl3 ---> 3AgCl + Fe(NO3)3

8. CaCO3 + 2HCl ---> CaCl2 + H2O + CO2

9. 4P + 4O2 ---> 2P2O5

10. 2Fe + 3Cl2 ---> 2FeCl3

giúp mình , mình cần gấp

\(a,4Na+O_2\xrightarrow{t^o}2Na_2O\\ b,3Fe+2O_2\xrightarrow{t^o}Fe_3O_4\\ c,2Al+6HCl\to 2AlCl_3+3H_2\\ d,2NaOH+H_2SO_4\to Na_2SO_4+2H_2O\\ e,2Ba+O_2\xrightarrow{t^o}2BaO\\ f,2KClO_3\xrightarrow[MnO_2]{t^o}2KCl+3O_2\\ g,3Fe+2O_2\xrightarrow{t^o}Fe_3O_4\\ h,2Al+3CuSO_4\to Al_2(SO_4)_3+3Cu\)

a: 4Al+3O2->2Al2O3

b: Mg+CuSO4->MgSO4+Cu

c: Cu(OH)2+2HCl->CuCl2+2H2O

d: 2Fe(OH)3->Fe2O3+3H2O

a)3Al+3Cl₂ =>2AlCl₃

b)3Fe+O₂=>Fe₃O₄

_{.................}

a) \(2Al+3Cl_2\rightarrow2AlCl_3\)

b) \(3Fe+2O_2\rightarrow Fe_3O_4\)

c) \(CuO+2HCl\rightarrow CuCl_2+H_2O\)

d) \(2Mg+O_2\rightarrow2MgO\)

e) \(Mg+2HCl\rightarrow MgCl_2+H_2\)

g) \(Zn+CuSO_4\rightarrow ZnSO_4+Cu\)

h) \(2Zn+O_2\rightarrow2ZnO\)

k) Giống câu e nha, lặp đề

m) \(Fe+CuSO_4\rightarrow FeSO_4+Cu\)

n) \(2KClO_3\rightarrow2KCl+3O_2\)

p) \(O_2+2H_2\rightarrow2H_2O\)

PTHH:

a) 2Al + 3Cl\(_2\) \(\underrightarrow{to}\) 2AlCl\(_3\)

b) 3Fe + 2O\(_2\) \(\underrightarrow{to}\) Fe\(_3\)O\(_4\)

c) CuO + 2HCl ---> CuCl\(_2\) + H\(_2\)O

d) 2Mg + O\(_2\) \(\underrightarrow{to}\) 2MgO

e) Mg + 2HCl ---> MgCl\(_2\) + H\(_2\)\(\uparrow\)

g) Zn + CuSO\(_4\) ---> ZnSO\(_4\) + Cu\(\downarrow\)

h) 2Zn + O\(_2\) \(\underrightarrow{to}\) 2ZnO

k) Mg + 2HCl ---> MgCl\(_2\) + H\(_2\)\(\uparrow\)

m) Fe + CuSO\(_4\) ---> FeSO\(_4\) + Cu\(\downarrow\)

n) 2KClO\(_3\) \(\underrightarrow{to}\) 2KCl + 3O\(_2\)\(\uparrow\)

P) O\(_2\) + 2H\(_2\) \(\underrightarrow{to}\) 2H\(_2\)O

Gọi CTTQ là :Fe_xO_y

Ta có:

\(\%Fe=\dfrac{NTK_{Fe}.x.100\%}{PTK_{FexOy}}\)

\(\Leftrightarrow70=\dfrac{56x.100}{56x+16y}\)

\(\Leftrightarrow3920x+1120y=5600x\)

\(\Rightarrow1120y=1680x\)

\(\Rightarrow\dfrac{x}{y}=\dfrac{1120}{1680}=\dfrac{2}{3}\Rightarrow x=2,y=3\)

CTN: (Fe₂O₃)_n=160

=> n=1

Vậy CTHH là : Fe₂O₃

CHÚC BẠN HỌC TỐT!!

Câu 1: a) +) \(FeO\)\(\Rightarrow\%Fe=\dfrac{56}{56+16}.100\%\approx77,78\%\)

+) \(Fe_2O_3\Rightarrow\%Fe=\dfrac{2.56}{2.56+3.16}.100\%=70\%\)

+) \(Fe_3O_4\Rightarrow\%Fe=\dfrac{3.56}{3.56+4.16}.100\%\approx72,41\%\)

+) \(Fe\left(OH\right)_2\Rightarrow\%Fe=\dfrac{56}{56+\left(16+1\right).2}.100\%\approx62,22\%\)

+) \(Fe\left(OH\right)_3\Rightarrow\%Fe=\dfrac{56}{56+\left(16+1\right).3}.100\%\approx52,34\%\)

+) \(Fe_2\left(SO_4\right)_3\Rightarrow\%Fe=\dfrac{2.56}{2.56+\left(32+4.16\right).3}.100\%=28\%\)

+) \(FeSO_4.7H_2O\Rightarrow\%Fe=\dfrac{56}{\left(56+32+4.16\right)+7.\left(2.1+16\right)}.100\%\approx20,14\%\)b) +) \(CO\Rightarrow\%C=\dfrac{12}{12+16}.100\%\approx42,96\%\)

+) \(CO_2\Rightarrow\%C=\dfrac{12}{12+2.16}.100\%\approx27,27\%\)

+) \(H_2CO_3\Rightarrow\%C=\dfrac{12}{2.1+12+3.16}.100\%\approx19,35\%\)

+) \(Na_2CO_3\Rightarrow\%C=\dfrac{12}{2.23+12+3.16}.100\%\approx11,32\%\)

+) \(CaCO_3\Rightarrow\%C=\dfrac{12}{40+12+3.16}.100\%=12\%\)

+) \(Mg\left(HCO_3\right)_2\Rightarrow\%C=\dfrac{2.12}{24+\left(1+12+3.16\right).2}.100\%\approx16,44\%\)

Câu cuối bài 2 hình như đề nhầm?

Câu 2: +) \(\left\{{}\begin{matrix}m_{CuO}=0,23.\left(64+16\right)=18,4\left(g\right)\\\%Cu=\dfrac{64}{64+16}.100\%=80\%\end{matrix}\right.\)

\(\Rightarrow m_{Cu}=18,4.80\%=14,72\left(g\right)\)

+) \(\left\{{}\begin{matrix}m_{CuSO4}=12\left(g\right)\\\%Cu=\dfrac{64}{64+32+4.16}.100\%=40\%\end{matrix}\right.\)

\(\Rightarrow m_{Cu}=12.40\%=4,8\left(g\right)\)

+) \(\left\{{}\begin{matrix}n_{CuSO4}=\dfrac{5}{\left(64+32+4.16\right)+5\left(2.1+16\right)}=0,02\left(mol\right)\\\%Cu=\dfrac{64}{64+32+4.16}.100\%=40\%\end{matrix}\right.\)

\(\Rightarrow m_{Cu}=\left(0,02.\left(64+32+4.16\right)\right).40\%=1,28\left(g\right)\)

a, C + O₂ -> CO2 (phản ứng hóa hợp)

b, 2KClO₃ -> 2KCl + 3O2 (phản ứng phân hủy)

c, H₂ + O₂ -> H₂O (phản ứng hóa hợp)

d, H₂ + CuO -> Cu + H₂O (phản ứng thế)

e, Fe + CuSO₄ -> FeSO4 + Cu (phản ứng thế)

f, 2Al + 6HCl -> 2AlCl₃ + 3H₂ (phản ứng thế)