Bài 19: Luyện tập về phản ứng oxi hóa khử

Câu 1:

a) 4Al + 3O₂ --t^o--> 2Al₂O₃

b) 2Fe + 6H₂SO₄ --> Fe₂(SO₄)₃ + 3SO₂ +6 H₂O

c) Fe₃O₄ + 10HNO₃ --> 3Fe(NO₃)₃ + NO₂ + 5H₂O

d) \(10Al+38HNO_3->10Al\left(NO_3\right)_3+2NO+3N_2O+19H_2O\)

\(\dfrac{30.n_{NO}+44.n_{N_2O}}{n_{NO}+n_{N_2O}}=19,2.2=38,4=>\dfrac{n_{NO}}{n_{N_2O}}=\dfrac{2}{3}\)

e) \(\left(5x-2y\right)M+\left(6nx-2ny\right)HNO_3->\left(5x-2y\right)M\left(NO_3\right)_n+nN_xO_y+\left(3nx-ny\right)H_2O\)

Đặt \(n_{Al}=x(mol);n_{Fe}=y(mol)\)

\(\Rightarrow 27x+56y=20,85(1)\\ n_{H_2}=\dfrac{11,76}{22,4}=0,525(mol)\\ 2Al+6HCl\to 2AlCl_3+3H_2\\ Fe+2HCl\to FeCl_2+H_2\\ \Rightarrow 1,5x+y=0,525(2)\\ (1)(2)\Rightarrow x=0,15(mol);y=0,3(mol)\\ a.\begin{cases} \%_{Al}=\dfrac{0,15.27}{20,85}.100\%=19,42\%\\ \%_{Fe}=100\%-19,42\%=80,58\% \end{cases}\)

\(b.\Sigma n_{HCl}=3x+2y=1,05(mol)\\ \Rightarrow m_{dd_{HCl}}=\dfrac{1,05.36,5}{20\%}=191,625(g)\\ c,n_{AlCl_3}=x=0,15(mol);n_{FeCl_2}=y=0,3(mol)\\ \Rightarrow m_{muối}=0,15.133,5+0,3.127=58,125(g)\)

2Al + 6HCl --> 2AlCl₃ + 3H₂

2NH₃ + 3Cl₂ --t^o--> N₂ + 6HCl 

Chất khử: NH₃, chất oxh: Cl₂

\(n_{Al}=\dfrac{6,21}{27}=0,23\left(mol\right)\)

Gọi số mol NO là a => số mol N₂O là 2,5a

Al⁰ - 3e --> Al⁺³

0,23->0,69

58H⁺ + 12NO₃^- 46e --> 2NO + 5N₂O + 29H₂O

_____________23a<-----a---->2,5a

Bảo toàn e: 23a = 0,69

=> a = 0,03 (mol)

=> n_NO = 0,03(mol); n_N2O = 2,5.0,03 = 0,075 (mol)

=> V_khí = (0,03+0,075).22,4 = 2,352 (l)

\(n_{Cu}=0,2(mol)\\ n_{NO}=a(mol);n_{NO_2}=b(mol)\\ Cu\to Cu^{2+}+2e\\ 0,2..................0,4\\ N^{+5}+1e\to N^{+4}\\ .........b.....b\\ N^{+5}+3e\to N^{+2}\\ ........3a.....a\)

Bảo toàn e: \(3a+b=0,4(1)\)

\(d_{khí/H_2}=19\\ \Rightarrow \dfrac{30a+46b}{a+b}=2.19\\ \Rightarrow 30a+46b=38(a+b)\\ \Rightarrow 4a-4b=0\Rightarrow a=b(2)\\ (1)(2)\Rightarrow a=b=0,1(mol)\\ \Rightarrow V_{NO}=V_{NO_2}=0,1.22,4=2,24(l)\)

Gọi số mol Fe, Cu là a, b 

=> 56a + 64b = 1,84

Fe⁰-3e-->Fe⁺³

a->3a

Cu⁰-2e-->Cu⁺²

b->2b

N⁺⁵ +3e--> N⁺²

___0,03<-0,01

N⁺⁵ +1e--> N⁺⁴

___0,04<-0,04

Bảo toàn e: 3a + 2b = 0,07

=> a = 0,01 ; b = 0,02

=> \(\left\{{}\begin{matrix}\%Fe=\dfrac{56.0,01}{1,84}.100\%=30,435\%\\\%Cu=\dfrac{0,02.64}{1,84}.100\%=69,565\%\end{matrix}\right.\)