14, mấy % á em?

`Fe + 2HCl -> FeCl_2 + H_2`

`2Na + 2HCl -> 2NaCl + H_2`

`2Al + 6HCl -> 2AlCl_3 + 3H_2`

`2Al(OH)_3 + 3H_2SO_4 -> Al_2(SO_4)_3 + 6H_2O`

Fe + 2HCl → FeCl₂ + H₂
2Na + 2HCl → 2NaCl + H₂
2Al + 6HCl → 2AlCl₃ + 3H₂
2Al + 3H₂SO₄ → Al₂(SO₄)₃ + 6H₂O

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

\(2Na+2HCl\rightarrow2NaCl+H_2\)

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

\(2Al\left(OH\right)_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+6H_2O\)

1. \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)

\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

2. Gọi: \(\left\{{}\begin{matrix}n_{Zn}=x\left(mol\right)\\n_{Al}=y\left(mol\right)\end{matrix}\right.\) ⇒ 65x + 27y = 9,2 (1)

Theo PT: \(n_{H_2}=n_{Zn}+\dfrac{3}{2}n_{Al}=x+\dfrac{3}{2}y=\dfrac{5,6}{22,4}=0,25\left(mol\right)\left(2\right)\)

Từ (1) và (2) ⇒ x = y = 0,1 (mol)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Zn}=\dfrac{0,1.65}{9,2}.100\%\approx70,65\%\\\%m_{Al}\approx29,35\%\end{matrix}\right.\)

3. Theo PT: \(\left\{{}\begin{matrix}n_{ZnSO_4}=n_{Zn}=0,1\left(mol\right)\\n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{2}n_{Al}=0,05\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{ZnSO_4}=0,1.160=16\left(g\right)\\m_{Al_2\left(SO_4\right)_3}=0,05.342=17,1\left(g\right)\end{matrix}\right.\)

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

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

Bài 3:

2Al + 3H₂SO₄ → Al₂(SO₄)₃ + 3H₂

\(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)

a) Theo PT: \(n_{H_2SO_4}=\dfrac{3}{2}n_{Al}=\dfrac{3}{2}\times0,2=0,3\left(mol\right)\)

\(\Rightarrow m_{H_2SO_4}=0,3\times98=29,4\left(g\right)\)

b) Theo PT: \(n_{H_2}=\dfrac{3}{2}n_{Al}=\dfrac{3}{2}\times0,2=0,3\left(mol\right)\)

Số phân tử H₂ là: \(0,3\times6\times10^{23}=1,8\times10^{23}\left(phântử\right)\)

Ta có: \(n_H=2n_{H_2}=2\times0,3=0,6\left(mol\right)\)

Số nguyên tử H là: \(0,6\times6\times10^{23}=3,6\times10^{23}\left(nguyêntử\right)\)

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

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

a) 2 \(\rightarrow\)

b) 2 \(\rightarrow\)

a) \(K_2O+H_2O->2KOH\)

b) \(2Fe+3Cl_2->2FeCl_3\)

c) \(2Al\left(OH\right)_3+3H_2SO_4->Al_2\left(SO_4\right)_3+6H_2O\)

d) \(4FeS_2+11O_2->2Fe_2O_3+8SO_2\)

e)\(2Al+6HCl->2AlCl_3+3H_2\)

f) \(2KClO_3->KCl+3O_2\)

g) \(4P+5O_2->2P_2O_5\)

h) \(3Fe+2O_2->Fe_3O_4\)

i) \(2Al+3CuSO_4->Al_2\left(SO_4\right)_3+3Cu\)

j) \(2K+2H_2O->2KOH+H_2\)

1. \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)

\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

2. Gọi: \(\left\{{}\begin{matrix}n_{Zn}=x\left(mol\right)\\n_{Al}=y\left(mol\right)\end{matrix}\right.\) ⇒ 65x + 27y = 17,7 (1)

Theo PT: \(n_{H_2}=n_{Zn}+\dfrac{3}{2}n_{Al}=x+\dfrac{3}{2}y=\dfrac{5,6}{22,4}=0,25\left(mol\right)\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{66}{235}\\y=-\dfrac{29}{1410}\end{matrix}\right.\)

Tới đây thì ra số mol âm, bạn xem lại đề nhé.

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

nAl = 5,4 : 27 = 0,2 mol

Theo pt: nH₂ = \(\dfrac{3}{2}\)nAl = 0,3 mol

=> V H₂ = 0,3.22,4 = 6,72 lít

b)nAl₂(SO₄)₃ = \(\dfrac{1}{2}nAl=0,1mol\)

=>mAl₂(SO₄)₃ = 0,1.342 = 34,2g