what will be the theoretical yeild, in grams, of potassium(iii) oxalate obtainable from this reaction

a. The limiting reactant is K₂C₂O₄

b. Mass of K₃Fe(C₂O₄)₃.3H₂O =  3.29 g

Note: The question is incomplete. The complete question is found in the text and attachment below:

If these quantities are used to synthesize potassium iron (III) oxalate, which reactant will be the limiting reagent?

What will be the theoretical yield in grams of potassium iron (III) oxalate obtainable from this reaction?

Explanation:

Equation of the reaction is given below:

2Fe(NH₄)₂(SO₄)₂.6H₂O + 3H₂C₂O₄ + H₂O₂ + 3K₂C₂O₄ ---> 2K₃Fe(C₂O₄)₃.3H₂O + 4(NH₄)HSO₄ + 8H₂O

molar mass of potassium iron (III) oxalate, K₃Fe(C₂O₄)₃.3H₂O = 491 g/mol

molar mass of Fe(NH₄)₂(SO₄)₂.6H₂O = 392 g/mol

number of moles = mass/molar mass = molarity, M * volume (L)

Number of moles present in 5.0 g of Fe(NH₄)₂(SO₄)₂.6H₂O = 5.0/393 = 0.0128 moles

Number of moles present in 33.0 mL of 1.0 M H₂C₂O₄ = 33/1000 * 1 = 0.0330 moles

Number of moles present in 20.0 mL of 3% H₂O₂ (1.0 M) = 20.0/1000 * 1.0 = 0.0200 moles

Number of moles present in 10.0 mL saturated K₂C₂O₄ (2.0 M) = 10/1000 * 2.0 = 0.0100 moles

a. From the equation of reaction, mole ratio of the reactants are in the following order: 2 : 3 : 1 : 3

from the reacting masses and volumes given, mole ratio of reactants are in the order: 0.0128 : 0.0330 : 0.0200 : 0.0100

Therefore, the limiting reactant is K₂C₂O₄

b. Calculating theoretical yield using the limiting reactant

From the equation of reaction, 3 moles K₂C₂O₄ produces 2 moles K₃Fe(C₂O₄)₃.3H₂O

Therefore, 0.0100 moles K₂C₂O₄ will produce 2/3 *0.0100 moles K₃Fe(C₂O₄)₃.3H₂O = 0.0067 moles K₃Fe(C₂O₄)₃.3H₂O

mass of K₃Fe(C₂O₄)₃.3H₂O = number of moles * molar mass

mass of K₃Fe(C₂O₄)₃.3H₂O = 0.0067 * 491 = 3.29 g