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0.158 moles of KMnO4 are present in 25.0 grams of potassium permanganate, KMnO4.
no. of moles = mass (g) ÷ molar mass (g/mol)
The molar mass of KMnO4 = 39 + 55 + 16(4) = 158g/mol
no. of moles = 25g ÷ 158g/mol
no. of moles = 0.158mol
Therefore, 0.158 moles of KMnO4 are present in 25.0 grams of potassium permanganate, KMnO4.
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