Please I need help with this Balance the following chemical

Balance the following chemical equations K2Cro, __K:[Fe(CO.hl. 3110 Transfer the balanced chemical reactions to page 6. 4. Use the first of the above balanced chemical equations to determine the following a. The reaction mixture initially contains 2.5 g of ferrous ammonium sulfate hexahydrate, Determine the Fe( NHa);(SO4)2 · limiting reactant. 611:0. and 10.00 ml, or i,5 M oxalic acid, HC:0. b. Determine the theoretical yield in grams of iron(II) oxalate dihydrate, FeC:0 2H:O. Given an actual yield of 0.99 g iron(II) oxalate, Fec:04·2H20, calculate the percentage yield c. Page 5 o Lab 5-Synthesis & Analysis of a Coordination Compound: Part I-Preparation of Ko[Fe(C,04b) 3H:0 ry University Department of Chemistry

Solution

Ans. Balanced reaction:

I. Fe(NH₄)₂(SO₄)₂.6H₂O + H₂C₂O₄ -------->

FeC₂O₄.2H₂O + (NH₄)₂SO₄ + H₂SO₄ + 4 H₂O

II. 2 FeC₂O₄.2H₂O + H₂C₂O₄ + H₂O₂ + 3 K₂C₂O₄ ------> 2 K₃[Fe(C₂O₄)₃].3H₂O

Note that reaction II has 2 H₂O molecules excess on the product side because K₂C₂O₄ is taken as anhydrous form instead of its monohydrate form K₂C₂O₄.H₂O.

The actual reaction taking place is-

2 FeC₂O₄.2H₂O + H₂C₂O₄ + H₂O₂ + 3 K₂C₂O₄.H₂O ------> 2 K₃[Fe(C₂O₄)₃].3H₂O

#4a. Theoretical molar ratio of reactants for reaction I:

                        Fe(NH₄)₂(SO₄)₂.6H₂O : H₂C₂O₄ = 1 : 1

# Moles of Fe(NH₄)₂(SO₄)₂.6H₂O taken = Mass / Molar mass

                                                = 2.5 g / (392.14288 g/ mol)

                                                = 0.006375 mol

# Moles of H₂C₂O₄ taken = Molarity x Volume of solution in liters

                                                = 1.5 M x 0.010 L

                                                = 0.015 mol

# Experimental molar ratio of reactants:

Fe(NH₄)₂(SO₄)₂.6H₂O : H₂C₂O₄ = 0.006375 mol : 0.015 mol = 1 : 2.35

Since the experimental moles of oxalic acid is greater than its theoretical value of 1 mol while keeping the moles of Fe(NH₄)₂(SO₄)₂.6H₂O constant at 1 mol, oxalic acid is the reagent in excess.

So,

            The limiting reactant is- Fe(NH₄)₂(SO₄)₂.6H₂O

#4b. Following stoichiometry, 1 mol Fe(NH₄)₂(SO₄)₂.6H₂O produces 1 mol FeC₂O₄.2H₂O.

So,

            Theoretical moles of FeC₂O₄.2H₂O formed = 0.006375 mol

            Theoretical mass of FeC₂O₄.2H₂O formed = 0.006375 mol x (179.89716 g/ mol)

                                                                                    = 1.1468 g

Therefore, theoretical yield of FeC₂O₄.2H₂O = 1.1468 g

#4c. % yield = (Actual yield / theoretical yield) x 100

                        = (0.99 g / 1.1468 g) x 100

                        = 86.33 %