Given: Formula units of #"Fe"_2"O"_3#; chemical equation (understood)
Find: Moles of #"O"_2#
Strategy:
The central part of any stoichiometry problem is to convert moles of something to moles of something else.
(a) We start with the balanced chemical equation for the reaction.
(b) We can use the molar ratio to convert moles of #"Fe"_2"O"_3# to moles of #"O"_2#.
#"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 #"Fe"_2"O"_3# to moles of #"Fe"_2"O"_3#.
Our complete strategy is:
#"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
#"2Fe"_2"O"_3 → "4Fe" + "3O"_2#
(b) #"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)#"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 #color(red)("0.999 mol O"_2)#.
