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First start with what you know to find the total moles.

Mn = 54.94g

O = 16g

54.94g * 2 = 109.88g

16 * 3 = 48g

Add these to get the weight of Mn2O3 which = 157.88

Divide 157.88 / 286 to get your total moles present, which = .552 Mols

To find moles Mn, divide 54.94 by the total weight which is 157.88. This gives you your moles Mn, which is .348 mols. Since you have Mn2, multiply that number by 2 to get .695 mols

Now you can divide Mol(Mn) by your total moles (.552) to get your moles Mn

So .695 mol Mn2O3 / .552 mol Mn = 1.1 Mol Mn

since 1.1 * 54.94 = 60 pct of the Molar Mass, this number makes sense since there is more Mn than O, the proportion seems correct.

hope this helped!

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13y ago
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5mo ago

To find the number of moles of Mn atoms in KMnO4, you first need to calculate the molar mass of KMnO4 (158.034 g/mol). Next, find the molar mass of Mn (54.938 g/mol) in KMnO4. Then, find the number of moles of Mn atoms by dividing the mass of KMnO4 by the molar mass of Mn.

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12y ago

16.1201 grams KMnO4 (1 mole KMnO4/158.04 grams)(1 mole Mn/1 mole KMnO4)(6.022 X 1023/1 mole Mn)(1 mole Mn atoms/6.022 X 1023)

= 0.102000 moles of manganese atoms

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Note that Avogadro's number is actually over itself in a form of one, so you do not need to go all the way in this, a formal set-up.

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The molar mass of manganese dioxide is 86,9368 g.

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3.00

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Q: How many moles of Mn atoms are there in 16.1201 grams of KMnO4?
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