How many grams are in 5.7 mol potassium permanganate?

1 Answer
anor277
Oct 31, 2016

There are about 900*g in a 5.7 molar quantity of KMnO_4.

Explanation:

The required mass is simply the product, "moles "xx" molar mass", i.e. 5.7*cancel(mol)xx158.034*g*cancel(mol^-1)=??*g.

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