0.24 moles of Fe2O3 will be produced from 27.0g of Fe assuming O2 is available in excess.
Explanation:
Data given:
mass of Fe = 27 grams
mass of oxygen = excess
moles of [tex]Fe_{2}[/tex][tex]O_{3}[/tex] = ?
Atomic mass of Fe= 55.48 grams/mole
atomic mass of Fe2O3 = 159.69 grams/mole
moles of Fe given =0.48 moles
number of moles = [tex]\frac{mass}{atomic mass of 1 mole}[/tex]
= [tex]\frac{27}{55.48}[/tex]
= 0.48 moles
Balanced chemical equation for the reaction is:
4 Fe + 3 [tex]O_{2}[/tex] ⇒ 2 [tex]Fe_{2}[/tex][tex]O_{3}[/tex]
4 moles of Fe gives 2 mole of Fe2O3
So, 0.48 moles will give x mole of Fe2O3
[tex]\frac{2}{4}[/tex] = [tex]\frac{x}{0.48}[/tex]
4x = 2 x 0.48
x = [tex]\frac{2 X 0.48}{4}[/tex]
x = 0.24 moles
so 0.24 moles of [tex]Fe_{2}[/tex][tex]O_{3}[/tex] is formed from 27 grams of Fe.
