# AP Chemistry

A 40.3 g of iron ore is treated as follows. The iron in the sample is all converted by a series of chemical reactions to Fe2O3. The mass of Fe2O3 is measured to be 10.5 grams. What was the percent iron in the sample of ore?

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1. Isn't this done the same as the problems before? Try it yourself first.

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2. Given:
m{sample} = 40.3g
m{Fe2O3} = 10.5g
w{Fe2O3} = 159.6887 ± 0.0002 g/mol
w{Fe} = 55.8452 ± 0.0001 g/mol
And the stoichiometry:
m{Fe}/w{Fe} = 3 m{Fe2O3}/w{Fe2O3}

The percentage of iron in the sample is:
m{Fe}/m{sample} = ...

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3. Typo correction:
m{Fe}/w{Fe} = 2 m{Fe2O3}/w{Fe2O3}

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4. Find the number of moles of Fe2O3 in 10.5g of Fe2O3:

10.5g*(1 mole/159.6882 g)=moles of Fe2O3

Convert moles of Fe2O3 into moles of Fe:

Moles of Fe2O3*(2 moles of Fe/1 mole of Fe2O3)=moles of Fe

How many g of Fe are in the moles of Fe calculated above?

Moles of Fe*(55.845g/1mole)=mass of Fe

(mass of Fe/40.3g of Fe ore)*100=% of Fe in original sample

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