# Iron can be extracted from iron(III) oxide by heating it with coke (carbon). The other product in the extraction reaction is carbon dioxide. What mass of iron is produced by reacting 550g of iron(III) oxide with 60g of coke?

Aug 2, 2017

I get $\text{271.959 g Fe}$ also. I would choose c) for the answer.

#### Explanation:

Balanced Equation

$\text{2Fe"_2"O"_3 + "3C}$$\rightarrow$${\text{4Fe + 3CO}}_{2}$

This is a limiting reactant stoichiometry question. The maximum amount of $\text{Fe}$ that can be produced is determined by the limiting reactant. We have to determine the amount of iron that can be produced by $\text{500}$ grams of iron(III) oxide, and $60$ grams of coke.

Mass of Iron Produced by Iron(III) oxide.

color(red)("Convert the mass of iron(III) oxide to moles by multiplying by the inverse of its molar mass, 159.687 g/mol."

color(blue)("Determine the moles of iron produced by multiplying the moles iron by the mole ratio between iron and iron(III) oxide":
$\text{4 mol Fe} :$$\text{2 mol Fe"_2"O"_3}$

$\textcolor{g r e e n}{\text{Determine the mass of iron produced by multiplying the moles iron by its molar mass}}$.

$550 \text{g Fe"_2"O"_3xxcolor(red)((1"mol Fe"_2)/(159.687"g Fe"_2"O"_3))xxcolor(blue)((4"mol Fe")/(2"mol Fe"_2"O"_3))xxcolor(green)((55.845"g Fe")/(1"mol Fe"))="385 g Fe}$

Mass of Iron Produced by Coke

Follow the same procedure as above, substituting $\text{C}$ for $\text{Fe"_2"O"_3}$, and use the mole ratio $4 \text{mol Fe:}$ $3 \text{mol C}$.

$60 \text{g C"xxcolor(red)((1"mol C")/(12.011"g C"))xxcolor(blue)((4"mol Fe")/(3"mol C"))xxcolor(green)((55.845"g Fe")/(1"mol Fe"))="372 g Fe}$

Coke is the limiting reactant because it produces less iron than iron(III) oxide.

I didn't round to the proper number of sig figs because it would give the wrong answer.