How many milliliters of 0.250 M KMnO4 are needed to react with 3.55 g of Iron(II) Sulfate, FeSO4?
Added by Daniel E.
Step 1
First, we need to find the moles of FeSO4. To do this, we'll use the molar mass of FeSO4, which is 151.91 g/mol. Moles of FeSO4 = (3.55 g) / (151.91 g/mol) = 0.02337 mol Show more…
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