Question

How many grams of iron are in 21.6 g of iron(III) oxide?

Answer 1

Approx. `15*g` of metal.

Explanation:

`"Moles of iron oxide, " Fe_2O_3` `=` `(21.6*g)/(159.69*g*mol)`

`=` `0.135*mol` with respect to the oxide.

But by the composition of the oxide, there are thus `2xx0.135*molxx55.8*g*mol^-1` `"iron metal"` `=` `??g`.

An extraordinary percentage of our budgets goes into rust prevention; `"iron(III) oxide"` is only part of the redox chemistry. Once you put up a bridge or a skyscraper, which is inevitably steel-based structure, it will begin to corrode.

Answer 2

There are 15.1 g of iron in the sample.

Explanation:

Step 1. Calculate the molar mass of `"Fe"_2"O"_3`

`"molar mass" = "2 × 55.84 g Fe + 3 × 16.00 g O" = "111.68 g Fe + 48.00 g O" = "159.68 g Fe"_2"O"_3`

This tells us that there are 111.68 g `"Fe"` in 159.68 g `"Fe"_2"O"_3`.

Step 2. Calculate the mass of `"Fe"`

∴ In 21.6 g `"Fe"_2"O"_3`,

`"Mass of Fe" =21.6 color(red)(cancel(color(black)("g Fe"_2"O"_3))) × "111.68 g Fe"/(159.68 color(red)(cancel(color(black)("g Fe"_2"O"_3)))) = "15.1 g Fe"`