a) 2Al+ 3H2SO4-> Al2(SO4)3+3H2

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

a) 2 \(\rightarrow\)

b) 2 \(\rightarrow\)

a) PTHH :\(Fe_2O_3+3H_2SO_4->Fe_2\left(SO_4\right)_3+3H_2O\)

b) Ta có :\(n_{Fe_2O_3}=\dfrac{48}{160}=0,3\left(mol\right);n_{H_2SO_4}=\dfrac{58,8}{98}=0,6\left(mol\right)\)

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

1______3

\(\dfrac{1}{0,3}< \dfrac{3}{0,6}\)

=>\(n_{H_2SO_4}dư=>n_{H_2SO_{4\left(dư\right)}}=0,9mol=>m_{H_2SO_4\left(dư\right)}=0,9.98=88,2\left(g\right)\)

=>

a. 4P + 5O₂ \(\underrightarrow{t^o}\) 2P₂O₅

b. 4Al + 3O₂ \(\underrightarrow{t^o}\) 2Al₂O₃

c. Fe + 2HCl \(\rightarrow\) FeCl₂ + H₂\(\uparrow\)

d. H₂ + CuO \(\underrightarrow{t^o}\) Cu + H₂O

e. 3CO + Fe₂O₃ \(\underrightarrow{t^o}\) 2Fe + 3CO₂\(\uparrow\)

f. Cu + 2H₂SO₄ \(\rightarrow\) CuSO₄ + SO₂\(\uparrow\) + 2H₂O

g. Fe + 4HNO₃ \(\rightarrow\) Fe(NO₃)₃ + NO\(\uparrow\) + 2H₂O

h. 2Al + 3H₂SO₄ \(\rightarrow\) Al₂(SO₄)₃ + 3H₂\(\uparrow\)

i. Ca(HCO₃)₂ \(\underrightarrow{t^o}\) CaCO₃ + CO₂\(\uparrow\) + H₂O

à, sorry nha, lo làm cái kia nên quên lun ><

a, b là phản ứng hoá hợp

c, d, h là phản ứng thế

i là phản ứng phân huỷ

Bài 1:

\(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)

\(n_{Al_2O_3}=\dfrac{m}{M}=\dfrac{20,4}{102}=0,2\left(mol\right)\)

Theo PTHH, \(n_{Al}=2n_{Al_2O_3}=2\cdot0,2=0,4\left(mol\right)\)

\(m_{Al}=n\cdot M=0,4\cdot27=10,8\left(g\right)\)

Theo PTHH, \(n_{O_2}=\dfrac{3}{2}n_{Al_2O_3}=\dfrac{3}{2}\cdot0,2=0,3\left(mol\right)\)

\(V_{O_2}=n\cdot22,4=0,3\cdot22,4=6,72\left(l\right)\)

Bài 1:

4Al + 3O₂ \(\underrightarrow{to}\) 2Al₂O₃

\(n_{Al_2O_3}=\dfrac{20,4}{102}=0,2\left(mol\right)\)

a) theo PT: \(n_{Al}=2n_{Al_2O_3}=2\times0,2=0,4\left(mol\right)\)

\(\Rightarrow m=m_{Al}=0,2\times27=5,4\left(g\right)\)

b) theo PT: \(n_{O_2}=\dfrac{3}{2}n_{Al_2O_3}=\dfrac{3}{2}\times0,2=0,3\left(mol\right)\)

\(\Rightarrow V_{O_2}=0,3\times22,4=6,72\left(l\right)\)

CuO + H₂SO₄ → CuSO₄ + H₂O (1)

Al₂O₃ + 3H₂SO₄ → Al₂(SO₄)₃ + 3H₂O (2)

Gọi \(x,y\) lần lượt là số mol của CuO và Al₂O₃

Theo PT1: \(n_{CuSO_4}=n_{CuO}=x\left(mol\right)\)

TheoPT2: \(n_{Al_2\left(SO_4\right)_3}=n_{Al_2O_3}=y\left(mol\right)\)

Ta có: \(\left\{{}\begin{matrix}160x+342y=57,9\\80x+102y=25,5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,255\\y=0,05\end{matrix}\right.\)

Vậy \(n_{CuO}=0,255\left(mol\right);n_{Al_2O_3}=0,05\left(mol\right)\)

\(\Rightarrow m_{Al_2O_3}=0,05\times102=5,1\left(g\right)\)

\(\Rightarrow\%Al_2O_3=\dfrac{5,1}{25,5}\times100\%=20\%\)

a, Cu+2H2SO4(đ)-t^o-->CuSO4+SO2+2H2O

b, Zn+2H2SO4(đ)-t^o-->ZnSO4+SO2+2H2O

2Fe+6H2SO4(đ)-t^o-->Fe2(SO4)3+3SO2+6H2O

Cu + 2H₂SO_{4 đặc} → CuSO₄ + 2H₂O + SO₂(t^o)

Ba + 2H₂SO_{4 đặc} → BaSO₄ + 2H₂O + SO₂(t^o)

2Al + 6H₂SO_{4 đặc} → Al(SO₄)₃ + 6H₂O + 3H₂O (t^o)

1) $2n+O_2\to 2nO$

2) $\mathrm{2F}e+\mathrm{3C}{l}_2\to \mathrm{2F}eC{l}_3$

3) $Mg+\mathrm{2H}Cl\to MgC{l}_2+H_2$

4) $\mathrm{2A}l+\mathrm{6H}Cl\to \mathrm{2A}lC{l}_3+{\mathrm{3H}}_2$

5) $\mathrm{2A}l+{\mathrm{3H}}_{2}S{O}_4\to A{l}_2{\left(S{O}_4\right)}_{{}_3}+{\mathrm{3H}}_2$

1:2n+O2 to→2nO

2: 2Fe+3Cl2 to→2FeCl3

3: Mg+2HCl to→MgCl2+H2

4: 2Al+6HCl →2AlCl3+3H2

5: 2Al+3H2SO4 →Al2(SO4)3+3H2