Preparation of tris(acetylacetonate)iron(III)
1. What is the reaction equation?
2. What are the Limiting Reagents used (please show full equation as to how go answer)?
3. What is the theoretical and actual yield of the final product?
4. What is the percent yield of the final product?
Final answer: The reaction equation for the preparation of tris(acetylacetonate)iron(III) is FeCl3•6H2O + 3(acacH) + 3NaOAc → Fe(acac)3 + 3NaCl + 6H2O. The limiting reagent in this reaction is FeCl3•6H2O. The theoretical yield of the final product is 0.5298 grams, and the actual yield is 0.0196 grams. Therefore, the percent yield of the final product is 3.70%.
The reaction equation for the preparation of tris(acetylacetonate)iron(III) is: FeCl3•6H2O + 3(acacH) + 3NaOAc → Fe(acac)3 + 3NaCl + 6H2O where FeCl3•6H2O is iron(III) chloride hexahydrate, acacH is 2,4 pentandione, and NaOAc is sodium acetate.
To determine the limiting reagents, we need to compare the moles of each reactant to the stoichiometric ratio in the balanced equation. The balanced equation shows that 1 mole of FeCl3•6H2O reacts with 3 moles of acacH and 3 moles of NaOAc to produce 1 mole of Fe(acac)3. Let's calculate the moles of each reactant: - FeCl3•6H2O: 0.4055 g / 270.2957 g/mol = 0.0015 mol - acacH: 5 mL * 0.600 mol/L = 0.003 mol - NaOAc: 2.5 mL * 0.450 mol/L = 0.0011 mol Based on the stoichiometric ratio, we can see that FeCl3•6H2O is the limiting reagent because it has the smallest number of moles.
The theoretical yield of the final product can be calculated using the moles of the limiting reagent. From the previous calculation, we know that the moles of FeCl3•6H2O is 0.0015 mol. The molar mass of Fe(acac)3 is 353.2 g/mol. Therefore, the theoretical yield is: Theoretical yield = 0.0015 mol * 353.2 g/mol = 0.5298 g The actual yield of the final product is given as 19.6 mg, which is equal to 0.0196 g.
The percent yield of the final product can be calculated using the formula: Percent yield = (actual yield / theoretical yield) * 100% Percent yield = (0.0196 g / 0.5298 g) * 100% = 3.70%