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How many moles are present in 25.0 grams of potassium
permanganate, KMnO4?

Sagot :

samueladesida43

0.158 moles of KMnO4 are present in 25.0 grams of potassium permanganate, KMnO4.

NUMBER OF MOLES:

  • The number of moles in a substance can be calculated by dividing the mass of the substance by its molar mass. That is;

no. of moles = mass (g) ÷ molar mass (g/mol)

  • According to this question, there are 25 grams of potassium permanganate (KMnO4).

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.

Learn more about number of moles at: https://brainly.com/question/14919968

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