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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