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jeyben [28]
3 years ago
5

How many moles are in 79.85g Fe2O3?

Chemistry
2 answers:
strojnjashka [21]3 years ago
8 0

Answer:

in 79.85 g of Fe2O3 there will be 0.5 moles of Fe2O3

Explanation:

To find the number of moles we must calculate the molecular weight of iron oxide, for that we use the periodic table:

mFe = 55.845 g/mol

mO = 15.999 g/mol

mFe2O3 = (2x55.845) + (3x15.999) = 159.687 g/mol

1 mole of iron oxide weighs 159.687 g, therefore, to calculate the number of moles present in 79.85 g it will be equal to:

nFe2O3 = 79.85/159.687 = 0.5 moles

mamaluj [8]3 years ago
4 0
<span>Find molar mass of Fe2O3 = 159.7 g/mol
Then: 79.85g/159.7 g/mol = 0.5 mol</span>
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