theoretical yield

Question

The theoretical yield of a reaction is the amount of product obtained if the limiting reactant is completely converted to product.

Consider the reaction:

4 Fe(s) + 3 O₂(g) → 2 Fe₂O₃(s)

If 6.220 g Fe is mixed with 15.97 g O₂, calculate the theoretical yield (g) of Fe₂O₃ produced by the reaction.

Answer 1

1) Find the molar masses of iron and oxygen gas:

ℳ_O₂ = 2·ℳ_O = 2·16 ^g/_mol = 32 ^g/_mol

ℳ_Fe = 55.845 ^g/_mol

2) Use these molar masses to find out how many moles of iron and oxygen gas you actually have:

n_O₂,actual = m_O₂,actual/ℳ_O₂ = 15.97 g/32 ^g/_mol = 0.4990625 mol (←keep extra figures to avoid rounding errors!!)

n_Fe,actual = m_Fe,actual/ℳ_Fe = 6.220 g/55.845 ^g/_mol = 0.111379712 mol

3) Determine the minimum amount of iron you would theoretically need to react all of the oxygen (and vice versa) using the stoichiometric ratios of the balanced equation:

n_{Fe,theoretical} = n_O₂,actual·^{4 Fe}/_{3 O₂} = 0.665416667 mol

n_{O₂,theoretical} = n_Fe,actual·^{3 O₂}/_{4 Fe} = 0.083534784 mol

4) The limiting reagent is that of which we have fewer moles than we theoretically need for a complete reaction:

n_Fe,actual < n_{Fe,theoretical} ⇒ Fe is the limiting reagent

5) Now that we know the limiting reagent, we can calculate the theoretical yield: the amount of product we will get given the amount of limiting reagent we actually have. This will, as in step 3, rely upon the stoichiometry of the balanced chemical equation:

n_{Fe₂O₃,theoretical} = n_Fe,actual·^{2 Fe₂O₃}/_{4 Fe} = 0.055689856 mol

6) Calculate the molar mass of Fe₂O₃:

ℳ_Fe₂O₃ = 2(55.845 ^g/_mol) + 3(16 ^g/_mol) = 159.69 ^g/_mol

7) Convert the theoretical yield of Fe₂O₃ to grams:

m_{Fe₂O₃,theoretical} = n_{Fe₂O₃,theoretical}·ℳ_Fe₂O₃ = 0.055689856 mol·159.69 ^g/_mol = 8.893 g

(note that the final answer is in the correct number of significant figures.)