a) KClO₃ + 6HCl --> KCl + 3Cl₂ + 3H₂O

Chất khử: HCl, chất oxh: KClO₃

b) 2KMnO₄ --t^o--> K₂MnO₄ + MnO₂ + O₂

Chất khử: KMnO₄, chất oxh: KMnO₄

\(a.QToxh:2\overset{-1}{Cl}\rightarrow Cl_2+2e|\times5\\QTkhử:2\overset{+5}{Cl}+10e\rightarrow \overset{0}{Cl_2}|\times1\)

HCl là chất oxi hóa, KClO₃ là chất khử

\(KClO_3+6HCl_{đặc}\rightarrow KCl+3Cl_2+3H_2O\)

\(b.QToxh:2\overset{-2}{O}\rightarrow\overset{0}{O_2}+4e|\times1\\ QTkhử:2\overset{+7}{Mn}+4e\rightarrow\overset{+6}{Mn}+\overset{+4}{Mn}|\times1\)

KMnO₄ vừa là chất khử, vừa là chất oxi hóa

\(2KMnO_4-^{t^o}\rightarrow K_2MnO_4+MnO_2+O_2\)

Gọi số mol Zn, Fe là a, b (mol)

=> 65a + 56b = 23,75 (1)

\(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)

PTHH: Zn + 2HCl --> ZnCl₂ + H₂

            a--------------->a------->a

            Fe + 2HCl --> FeCl₂ + H₂

            b----------------->b----->b

=> a + b = 0,4 (2)

(1)(2) => a = 0,15 (mol); b = 0,25 (mol)

=> m_Zn = 0,15.65 = 9,75 (g); m_Fe = 0,25.56 = 14 (g)

\(\left\{{}\begin{matrix}\%m_{Zn}=\dfrac{9,75}{23,75}.100\%=41,05\%\\\%m_{Fe}=\dfrac{14}{23,75}.100\%=58,95\%\end{matrix}\right.\)

b) m_ZnCl2 = 0,15.136 = 20,4 (g)

m_FeCl2 = 0,25.127 = 31,75 (g)

=> m_muối = 20,4 + 31,75 = 52,15 (g)

Gọi số mol Fe, Mg, Al, Zn là a,b,c,d (mol)

\(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\); \(n_{Cl_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

TN1:

Fe⁰ - 2e --> Fe⁺²

a--->2a

Mg⁰ - 2e --> Mg⁺²

b---->2b

Al⁰ - 3e --> Al⁺³

c---->3c

Zn⁰ - 2e --> Zn⁺²

d--->2d

2H⁺ + 2e --> H₂⁰

          0,2<--0,1

Bảo toàn e: 2a + 2b + 3c + 2d = 0,2 (1)

TN2:

Fe⁰ - 3e --> Fe⁺³

a--->3a

Mg⁰ - 2e --> Mg⁺²

b---->2b

Al⁰ - 3e --> Al⁺³

c---->3c

Zn⁰ - 2e --> Zn⁺²

d---->2d

Cl₂⁰ + 2e --> 2Cl^-

0,15->0,3

Bảo toàn e: 3a + 2b + 3c + 2d = 0,3 (2)

(1)(2) => a = 0,1 (mol)

=> \(\%m_{Fe}=\dfrac{0,1.56}{32}.100\%=17,5\%\)

66C

67D

68A

Gọi số mol Fe₂O₃, CuO là a, b (mol)

n_HCl = 0,3.2 = 0,6 (mol)

PTHH: Fe₂O₃ + 6HCl --> 2FeCl₃ + 3H₂O

                 a----->6a------->2a

            CuO + 2HCl --> CuCl₂ + H₂O

              b----->2b------->b

=> \(\left\{{}\begin{matrix}\dfrac{2a}{b}=\dfrac{3}{4}\\6a+2b=0,6\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}a=\dfrac{9}{170}\left(mol\right)\\b=\dfrac{12}{85}\left(mol\right)\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}\%m_{Fe_2O_3}=\dfrac{\dfrac{9}{170}.160}{\dfrac{9}{170}.160+\dfrac{12}{85}.80}.100\%=42,857\%\\\%m_{CuO}=\dfrac{\dfrac{12}{85}.80}{\dfrac{9}{170}.160+\dfrac{12}{85}.80}.100\%=57,143\%\end{matrix}\right.\)

59 C

60D

61D

B

b

C

C