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How many grams are in 1.70 moles of potassium permanganate \([tex]KMnO_4[/tex])? Please include solutions.

Answer :

Final answer:

The mass in grams of 1.70 moles of potassium permanganate (KMnO4) is calculated by multiplying the amount in moles (1.70 moles) by the molar mass of KMnO4 (158.04 g/mol). The result is 268.67 grams.

Explanation:

In order to calculate the mass in grams of 1.70 moles of potassium permanganate (KMnO4), we need to use the molar mass of KMnO4. The atomic masses of each of the atoms in KMnO4 are approximately: potassium (K) is 39.1 g/mol, manganese (Mn) is 54.94 g/mol, and oxygen (O) is 16 g/mol. Therefore, the molar mass of KMnO4 (the sum of all these atomic masses) is approximately 158.04 g/mol.

To calculate the mass in grams of 1.70 moles of KMnO4, we multiply the number of moles by the molar mass: 1.70 moles * 158.04 g/mol = 268.67 g KMnO4.

So, there are 268.67 grams in 1.70 moles of potassium permanganate.

Learn more about Molar Mass here:

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