ΔfH298=ΣΔ fH298(sp)  - ΣΔfH298 (cd) =  2.(-393,5) + 3.(-285,84) - (-84,7) = -1559,82 kJ.\(\)

v = ΔC\Δt = 0,22 - 0,1 \ 4 = 0,03 (M\s)

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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