1. 185 grams of iron (II) oxide were reacted with 38.0 grams of aluminum as follows: 3 FeO(s) + 2 Al(s) → Al2O3(s) + 3 Fe(g) a) What was the limiting reactant? b) How many grams of total product (Al2O3 + Fe) formed if the reaction goes to completion? c) The total mass of the reactants is 223 grams. However, this is not the total mass of the products. Why not? d) If the actual yield of the iron is 105.0 grams, what is the percent yield of iron?

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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