Answered

Welcome to Westonci.ca, your one-stop destination for finding answers to all your questions. Join our expert community now! Get detailed answers to your questions from a community of experts dedicated to providing accurate information. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals.

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

Thanks for using our platform. We're always here to provide accurate and up-to-date answers to all your queries. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Discover more at Westonci.ca. Return for the latest expert answers and updates on various topics.