a) The limiting reactant is aluminium (Al).
b) mass of the products = 406.36 g
c) Because the mass of the components at the end of the reaction is equal to the mass of the products plus the mass of unreacted iron (II) oxide.
d) percent yield of iron = 88.44 %
Explanation:
We have the chemical reaction:
3 FeO (s) + 2 Al (s) → Al₂O₃ (s) + 3 Fe (s)
a) First we calculate the number of moles of iron (II) oxide (FeO) and aluminium (Al):
number of moles = mass / molar weight
number of moles of FeO = 185 / 72 = 2.57 moles
number of moles of Al = 38 / 27 = 1.41 moles
From the chemical reaction we see that if 3 moles of FeO will react with 2 moles of Al then 2.57 moles of FeO will react with 1.71 moles of Al but we only have 1.41 moles of Al, so the limiting reactant will be Al.
b) Knowing the chemical reaction and the limiting reactant, we devise the following reasoning:
if 2 moles of Al produces 1 mole of Al₂O₃ and 3 moles of Fe
then 1.41 moles of Al produces X mole of Al₂O₃ and Y moles of Fe
X = (1.41 × 1) / 2 = 2.82 moles of Al₂O₃
Y = (1.41 × 3) / 2 = 2.12 moles of Fe
mass = number of moles × molar weight
mass of Al₂O₃ = 2.82 × 102 = 287.64 g
mass of Fe = 2.12 × 56 = 118.72 g (theoretical quantity, see point d.)
mass of the products = mass of Al₂O₃ + mass of Fe
mass of the products = 287.64 + 118.72 = 406.36 g
c) Because aluminium is the limiting reactant, some of the iron (II) oxide will remain unreacted and will be found at the end of the reaction. So the mass of the components at the end of the reaction is the mass of the products plus the mass of unreacted iron (II) oxide.
d) The formula for percent yield is:
percent yield = (practical quantity / theoretical quantity) × 100
percent yield of iron = (105 / 118.72) × 100 = 88.44 %
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