Please find the balanced equation, theoretical yield, atom economy and estimated e factor ) based on the procedure To prepare the catalyst, sodium molybdate dihydrate (0.30 g; 1.2 mmol), and

Question

Please find the balanced equation, theoretical yield, atom economy and estimated e factor )

based on the procedure

To prepare the catalyst, sodium molybdate dihydrate (0.30 g; 1.2 mmol), and 4 M HCl (0.5 mL; 2.0 mmol) were added to a vial. About 1 mL of water was added to complete dissolution. Into a second vial benzyl tri-ethyl ammonium chloride (BTEAC) (0.525 g; 2.30 mmol) and ca.3 mL water were stirred until dissolved.

Answer 1

8 Na₂MoO₄ + 12 HCl + 4 BnEt₃N⁺ Cl^- ---> [BnEt₃N]₄ Mo₈O₂₆ + 16 NaCl + 6 H₂O ... balanced equation

mmols Na₂MoO₄ = 1.2 mmol (÷8 ->0.15) = LIMITING REACTANT

mmols HCl = 2.0 mmol (÷12 -> 0.17)

mmols BnEt₃N⁺ Cl^- = 2.30 mmol (÷4 -> 0.58)

Theoretical yield = 1.2 mmol Na₂MoO₄ x 1 mmol [BnEt₃N]₄ Mo₈O₂₆ / 8 mmol Na₂MoO₄ = 0.15 mmol [BnEt₃N]₄ Mo₈O₂₆

To find the atom economy we look at the molar mass of the product divided by ∑ of molar masses of all the products. I'll leave it to you to find the molar masses of all the products, and then carry out this simple division:

MW [BnEt₃N]₄ Mo₈O₂₆ / MW NaCl + MW H₂O + MW [BnEt₃N]₄ Mo₈O₂₆

e factor is calculated as mass of waste / mass of desired product

The mass of the waste will be mass of NaCl plus mass of H2O produced. You can calculate that from the 1.2 mmols Na₂MoO₄ and the stoichiometry of the balanced equation. The mass of the desired product, [BnEt₃N]₄ Mo₈O₂₆ will be 0.15 mmols x the molar mass of [BnEt₃N]₄ Mo₈O₂₆ in mg (not grams).

Good luck.