What is the theoretical yield in grams of the precipitate starting with 1.250 moles of iron ions (charge is +3) and 60.00 grams of hydroxide ions?
Fe^+3 + 3OH^- ==> Fe(OH)3(s)
This is a limiting reagent problem.
Convert 60.00 g OH^- to mols.
a. Determine mols Fe(OH)3 produced by using 1.25 mols Fe^+3.
b. Determine mols Fe(OH)3 produced by using 60/17 mols OH^-.
c. The correct answer will be the smaller of the two answers in a or b. That material is the limiting reagent. Then convert the mols Fe(OH^-)3 produced to grams. That is the theoretical yield.
Then convert mols
To determine the theoretical yield of the precipitate, you need to identify the balanced equation for the reaction that occurs between the iron ions (Fe^3+) and hydroxide ions (OH^−). The balanced equation will allow you to determine the molar ratio between the reactants and products.
Step 1: Write the balanced equation for the reaction between iron ions and hydroxide ions.
Fe^3+ + 3OH^− → Fe(OH)3
The balanced equation shows that for every 1 mole of Fe^3+, 3 moles of OH^− are required to produce 1 mole of Fe(OH)3.
Step 2: Calculate the moles of hydroxide ions.
Given: 60.00 grams of hydroxide ions (OH^−)
To calculate the moles of OH^−, you need to use its molar mass, which is calculated by summing the atomic masses of the constituent atoms.
Molar mass of OH^−: (1 × atomic mass of O) + (1 × atomic mass of H) = (1 × 16.00) + (1 × 1.01) = 17.01 g/mol
Moles of OH^− = Mass of OH^− / Molar mass of OH^−
= 60.00 g / 17.01 g/mol
≈ 3.525 moles
Step 3: Determine the limiting reactant.
To determine the limiting reactant, compare the moles of Fe^3+ and OH^− using their molar ratio from the balanced equation.
From the balanced equation, 1 mole of Fe^3+ requires 3 moles of OH^− to produce Fe(OH)3.
Moles of OH^− available: 3.525 moles
Using the molar ratio, the number of moles of Fe^3+ required is calculated as:
Moles of Fe^3+ required = (3.525 moles OH^−) / (3 moles OH^−/1 mole Fe^3+)
≈ 1.175 moles
The calculation shows that 1.175 moles of Fe^3+ are required based on the available moles of OH^−. Since the starting amount of Fe^3+ is 1.250 moles, it is in excess. Therefore, OH^− is the limiting reactant.
Step 4: Calculate the theoretical yield of the precipitate.
The theoretical yield is the maximum amount of product that can be obtained assuming complete reaction of the limiting reactant.
Using the limiting reactant (OH^−), we can calculate the moles of Fe(OH)3 formed.
From the balanced equation, 1 mole of Fe(OH)3 is produced for every 3 moles of OH^−.
Moles of Fe(OH)3 = (3.525 moles OH^−) / (3 moles OH^−/1 mole Fe(OH)3)
≈ 1.175 moles
Step 5: Convert moles to grams.
To convert moles of Fe(OH)3 to grams, use its molar mass.
Molar mass of Fe(OH)3: (1 × atomic mass of Fe) + (3 × atomic mass of O) + (3 × atomic mass of H)
= (1 × 55.85) + (3 × 16.00) + (3 × 1.01)
≈ 106.85 g/mol
Theoretical yield = Moles of Fe(OH)3 × Molar mass of Fe(OH)3
≈ 1.175 moles × 106.85 g/mol
Finally, calculate the theoretical yield:
Theoretical yield ≈ 125.54 grams of Fe(OH)3