1. S+O₂->SO₂(t0)

2. 2SO₂+O₂->2SO₃(t0)

3. SO₃+H₂O->H₂SO₄

4. H₂SO₄+MgO->MgSO₄+H₂O

a,4FeS₂ + 11O₂ --->2 Fe₂O₃+ 4SO₂
2SO₂ +O₂ --->2SO₃
SO₃ + H₂O ---> H₂SO₄
 

Na2CO3(x) + 2HCl ---> 2NaCl + CO2(x ) +H2O

K2CO3(y ) + 2HCl ---> 2KCl + CO2(y ) +H2O

Đặt nNa2CO3 = x (mol); nK2CO3 = y (mol)

=> 106x + 138y = 38,2 (1)

nCO2 = 0,3 (mol)

=> x + y = 0,3 (mol) (2)

Từ (1,2) => x = 0,1 (mol) , y = 0,2 (mol)

=> % khối lượng

Theo PTHH: nHCl = 2nCO2 = 0,6 (mol)

\(\Rightarrow m_{ddHCl}=\dfrac{0,6.36,5.100}{10}=219\left(g\right)\)

Bài 1:

+ Oxit axit

SiO2:Silic đioxit

SO2: Lưu huỳnh đioxit

NO: Nito oxit

+ Oxit bazo

Fe2O3: Sắt (III) oxit

Cu2O: Đồng (I) oxit

Ag2O: Bạc(I) oxit

Bài 2:

a/ 2KMnO₄ → K₂MnO₄ + MnO₂ + O₂ → Phản ứng phân hủy

b/ Na₂O + H₂O → 2NaOH → Phản ứng hóa hợp

c/ 2Al + 3Cl₂ → 2AlCl₃ → Phản ứng hóa hợp

d/ 2Fe(OH)₃ → Fe₂O₃ + 3H₂O → Phản ứng phân hủy

Bài 3:

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

2,25___1,5___________

\(n_{Fe}=\frac{126}{56}=2,25\left(mol\right)\)

\(\Rightarrow V_{O2}=1,5.22,4=33,6\left(l\right)\)

\(2KClO_3\rightarrow2KCl+3O_2\)

1_________________1,5

\(\Rightarrow m_{KClO3}=1.\left(39+35,5+16.3\right)=122,5\left(g\right)\)

a) 2H\(_2\)O \(\rightarrow\)\(^{điệnphân}\)2H\(_2\) + O\(_2\)

Mol: 0,4 \(\rightarrow\) 0,4 : 0,2

O\(_2\)+ 2CO \(\rightarrow\)\(^{t^0}\) 2CO\(_2\)

Mol: 0,2 : 0,4 \(\rightarrow\) 0,4

b) Ta có: m\(_{H_2O}\)=7,2(g)

=> n\(_{H_2O}\) = 7,2 : 18 = 0,4(mol)

V\(_{O_2}\) = 0,2 . 22,4 =4,48(l)

c) m\(_{CO_2}\)= 0,4 . 44= 17,6(g)

2/\(Si+O_2\rightarrow SiO_2\)

\(2NaOH+SiO_2\rightarrow H_2O+Na_2SiO_3\)

\(H_2O+Na_2SiO_3+CO_2\rightarrow Na_2CO_3+H_2SiO_3\)

\(H_2SiO_3\rightarrow H_2O+SiO_2\)

\(2C+SiO_2\rightarrow2CO+Si\)

Bài 2:

- Si + O2 → SiO2

- SiO2 + NaOH → Na2SiO3 + H2O

- Na2SiO3 + CO2 → SiO2 + Na2CO3

- 2SiO2 + 2C → 2CO +Si

nAl= \(\dfrac{4,5}{27}\simeq0,17\left(mol\right)\)

a) 2Al + 6HCl -> 2AlCl3 + 3H2

0,17 --------------> 0,17 ----> 0,255

b) VH2 = 0,255 . 22,4 = 5,712 (l)

c) mAlCl3 = 0,17 . 133,5 =22,695(g)

a, 2Al + 6HCl -> \(2AlCl_3\) + \(3H_2\)\(\uparrow\)

b, \(n_{Al}\) = \(\dfrac{4,5}{27}\) \(\approx\) 0,16 (mol)

=> \(n_{H_2}\) = \(\dfrac{3}{2}\) \(n_{Al}\) = \(\dfrac{3}{2}\). 0, 16 = 0,24 (mol)

=> \(V_{H_2}\)dktc = 0,24 . 22,4 = 5,376 (l)

c, \(n_{AlCl_3}\) = \(n_{Al}\)= 0,16 mol

=> \(m_{AlCl_3}\) = 0,16 . ( 27 + 35,5 . 3) = 21,36 (g)

nAl=4,5/27~0,167(mol)

2Al+6HCl--->2AlCl3+3H2

0,167________0,167___0,2505

VH2=0,2505.22,4=5,6112(l)

mAlCl3=0,167.133,5=22,2945(g)

Đặt tỉ lệ

Rồi lập hệ pt

\(\text{a) }n_{H_2O}=\dfrac{m}{M}=\dfrac{0,9}{18}=0,05\left(mol\right)\)

\(pthh:CuO+H_2\overset{t^o}{\rightarrow}Cu+H_2O\left(1\right)\\ \text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }x\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }x\\ PbO+H_2\overset{t^o}{\rightarrow}Pb+H_2O\left(2\right)\\ \text{ }\text{ }\text{ }y\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }y\)

Từ (1) và (2), ta có hệ phương trình:

\(\left\{{}\begin{matrix}x+y=0,05\\80x+217y=8,59\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{113}{6850}\\y=\dfrac{459}{13700}\end{matrix}\right.\)

\(\Rightarrow m_{CuO}=n\cdot M=80\cdot x=80\cdot\dfrac{113}{6850}=1,32\left(g\right)\\ \Rightarrow m_{PbO}=n\cdot M=217\cdot\dfrac{459}{13700}=7,27\left(g\right)\)

\(\Rightarrow\%CuO=\dfrac{1,32\cdot100}{8,59}=15,37\%\\ \%PbO=\dfrac{7,27\cdot100}{8,59}=84,63\%\)

b) Theo \(pthh\left(1\right):n_{Cu}=n_{CuO}=\dfrac{113}{6850}\left(mol\right)\)

Theo \(pthh\left(2\right):n_{Pb}=n_{PbO}=\dfrac{459}{13700}\left(mol\right)\)

\(\Rightarrow m_{Cu}=n\cdot M=\dfrac{113}{6850}\cdot64=1,06\left(g\right)\\ m_{Pb}=n\cdot M=\dfrac{459}{13700}\cdot207=6,94\left(g\right)\\ \Rightarrow m_{h^2}=1,06+6,94=8\left(g\right)\)

\(\Rightarrow\%Cu=\dfrac{1,06\cdot100}{8}=13,25\%\\ \%Pb=\dfrac{6,94\cdot100}{8}=86,75\%\\ \)