Câu 3: đenta fH298 = đen ta fH298(sp) - đenta fH298(cd)

= 2.(-393,5) + 3.(-285,84) - (-84,7)

= -1559,82 kJ

Câu 2: v = đentaC \ đentat = 0,22 - 0,1 \ 4

= 0,03 (M\s)

Câu 1 :

a. PTHH 2KMnO4 + 16HClđặc ----> 2KCl + 2MnCl2 + 5Cl2 + 8H2O

Chất khử: HCl

Chất oxi hóa: KMnO4

Quá trình oxi hóa: 2Cl----> Cl20 + 2e |x5

Quá trình khử: Mn+7 + 5e ---> Mn+2 |x2

b, nNaI = 0,2.0,1 = 0,02 (mol)

PTHH: 2NaI + Cl2 ---> 2NaCl + I2

mol: 0,02 --> 0,01

=> PTHH: 2KMnO4 + 16HClđặc ----> 2KCl + 2MnCl2 + 5Cl2 + 8H2O

mol: 0,004 <--- 0,01

=> mKMnO4 = n.M = 0,004.158 =0,632(g)

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