Answer :
Given:98.1 mL of 5 M potassium hydroxide is mixed with 39.9 mL of 4.5 M Iron (III) acetate resulting in a precipitate of Iron (III) hydroxide.To calculate the theoretical yield in grams of Iron (III) hydroxide, the first step is to balance the chemical equation for the reaction that takes place between potassium hydroxide and iron (III) acetate. 3KOH + Fe(C2H3O2)3 → Fe(OH)3 + 3KC2H3O2The balanced chemical equation for the reaction that takes place between potassium hydroxide and iron (III) acetate can be represented as follows;3KOH + Fe(C2H3O2)3 → Fe(OH)3 + 3KC2H3O2The molar mass of Fe(OH)3 is calculated as follows;Molar mass of Fe(OH)3 = Atomic mass of Fe + (3 x Atomic mass of O) + (3 x Atomic mass of H) = (55.85 g/mol) + (3 x 16 g/mol) + (3 x 1 g/mol) = 106.85 g/molThus the molar mass of Fe(OH)3 is 106.85 g/mol.To determine the theoretical yield of Iron (III) hydroxide we must first determine the limiting reactant (the reactant that is fully consumed in the reaction) among potassium hydroxide and iron (III) acetate.Limiting ReactantIn order to find out the limiting reactant among potassium hydroxide and iron (III) acetate, we will first find out the number of moles of each using the formula;Moles = Concentration x Volume in Liters (L)Moles of KOH = Concentration of KOH × Volume of KOH = 5 M × (98.1 mL/1000 mL) = 0.4905 moles Moles of Fe(C2H3O2)3 = Concentration of Fe(C2H3O2)3 × Volume of Fe(C2H3O2)3 = 4.5 M × (39.9 mL/1000 mL) = 0.17955 molesBased on the balanced chemical equation, the mole ratio of KOH to Fe(C2H3O2)3 is 3:1. Hence, the limiting reactant is Fe(C2H3O2)3 since it is lesser in moles compared to KOH. This means that all of the 0.17955 moles of Fe(C2H3O2)3 will be consumed in the reaction while 0.4905 - (0.17955 x 3) = 0.05145 moles of KOH will be left over after the reaction is complete.The theoretical yield is then calculated using the limiting reactant. We can calculate the number of moles of Fe(OH)3 produced from 0.17955 moles of Fe(C2H3O2)3 using the balanced chemical equation. The mole ratio of Fe(C2H3O2)3 to Fe(OH)3 is 1:1. Hence;Moles of Fe(OH)3 = Moles of Fe(C2H3O2)3 = 0.17955 moles. The mass of Fe(OH)3 is then calculated using the formula;Mass = Number of moles × Molar massMass of Fe(OH)3 = Number of moles of Fe(OH)3 × Molar mass of Fe(OH)3 = 0.17955 moles × 106.85 g/mol = 19.179 gTherefore, the theoretical yield of Fe(OH)3 is 19.179 g.
The theoretical yield of iron (III) hydroxide is 19.19 grams.
What is the theoretical yield of iron (iii) hydroxide?
The balanced chemical equation for the reaction between potassium hydroxide and iron (III) acetate is:
3 KOH + Fe(C₂H₃O₂)₃ → Fe(OH)₃ + 3 KC₂H₃O₂
To calculate the theoretical yield of iron (iii) hydroxide, first, we determine the limiting reagent.
The number of moles of each reactant:
Number of moles (n) = Molarity (M) × Volume (V)
For potassium hydroxide (KOH):
n(KOH) = 5 M × 0.0981 L
number of moles = 0.4905 moles
For iron (III) acetate (Fe(C₂H₃O₂)₃):
number of moles = 4.5 M × 0.0399 L
number of moles = 0.17955 moles
Since the stoichiometric ratio is 1:1, the number of moles of Fe(OH)₃ = 0.17955 moles.
The molar mass of Fe(OH)₃ = 106.88 g/mol
Theoretical yield = Number of moles × Molar mass
Theoretical yield = 0.17955 moles × 106.88 g/mol
Theoretical yield= 19.19 grams
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