Determine what mass of potassium permanganate KMnO4 entered into the decomposition reaction if 0.25 mol of oxygen O2 was released.

univerkov | November 22, 2020 | education

1) Let’s compose the reaction equation: 2KMnO4 ==> K2MnO4 + MnO2 + O2 ^
2) According to the reaction equation, we find the amount of permanganate: v (KMnO4) = v (O2) * 2 = 0.25 mol * 2 = 0.5 mol
3) Find the mass of permanganate: m (KMnO4) = 0.5 mol * 158 g / mol = 79 g
Answer: m (KMnO4) = 79g

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