# How many moles of O_2 are produced from 4.01 * 10^23 molecules Fe_2O_3?

Jan 8, 2016

The reaction produces 0.999 mol of ${\text{O}}_{2}$.

#### Explanation:

Given: Formula units of ${\text{Fe"_2"O}}_{3}$; chemical equation (understood)

Find: Moles of ${\text{O}}_{2}$

Strategy:

The central part of any stoichiometry problem is to convert moles of something to moles of something else.

(b) We can use the molar ratio to convert moles of ${\text{Fe"_2"O}}_{3}$ to moles of ${\text{O}}_{2}$.

${\text{moles of Fe"_2"O"_3stackrelcolor (blue)("molar ratio"color(white)(Xl))( →) "moles of O}}_{2}$

(c) Then we can use Avogadro's number to convert formula units of ${\text{Fe"_2"O}}_{3}$ to moles of ${\text{Fe"_2"O}}_{3}$.

Our complete strategy is:

${\text{Formula units of Fe"_2"O"_3stackrelcolor (blue)("Avogadro's number"color(white)(Xl))(→) "moles of Fe"_2"O"_3stackrelcolor (blue)("molar ratio"color(white)(Xl))( →) "moles of O}}_{2}$

Solution

(a) The balanced equation is

${\text{2Fe"_2"O"_3 → "4Fe" + "3O}}_{2}$

(b) ${\text{Moles of Fe"_2"O"_3 = 4.01×10^23color(red)(cancel(color(black)("FU Fe"_2"O"_3))) × ("1 mol Fe"_2"O"_3)/(6.022×10^23color(red)(cancel(color(black)("FU Fe"_2"O"_3)))) = "0.6659 mol Fe"_2"O}}_{3}$

(c)${\text{Moles of O"_2 = 0.6659 color(red)(cancel(color(black)("mol Fe"_2"O"_3))) × ("3 mol O"_2)/(2 color(red)(cancel(color(black)("mol Fe"_2"O"_3)))) = "0.999 mol O}}_{2}$
(3 significant figures)

Answer: 4.01×10^23color(white)(l)"FU Fe"_2"O"_3 will produce $\textcolor{red}{{\text{0.999 mol O}}_{2}}$.