What mass of $"Fe"$ is in $"316.80 g Fe"_2"O"_3"$ ?

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Answer 1 · 2015-11-10T21:23:08

$"316.80 g Fe"_2"O"_3$ contains $"221.56 g Fe"$ .

One mole $"Fe"_2"O"_3"$ contains two moles $"Fe"$ atoms.

The molar mass of iron is 55.845 g/mol. Two moles of iron atoms equals $2xx55.845"g/mol Fe"="111.69 g/mol Fe"$ .

Determine the percentage of iron in one mole of iron(III) oxide, which has a molar mass of $"159.70 g/mol"$ .

$(111.69cancel"g/mol Fe")/(159.70cancel"g/mol Fe"_2"O"_3)xx100="69.937% Fe"$

In order to determine the grams of iron in 316.80 g of iron(III) oxide, multiply the percentage of $"Fe"$ times $316.80"g Fe"_2"O"_3"$ .

$0.69937 "Fe"xx316.80"g Fe"_2"O"_3="221.56 g Fe"$

What mass of "Fe""Fe" is in "316.80 g Fe"_2"O"_3""316.80 g Fe"_2"O"_3"?