Question

Vauh section. UL LUMPale di Conec For this lab you will once again have three separa once again have three separate sets of calculations: PARTI - Synthesis and purification of [Cu(NH3).J$0. i. limiting reagent determination (use the reactant) nt determination (use the stoichiometric ratio to determine the limiting CuSO4.5H20 (s) + 4NH: (aq) → [Cu(NH3)4]S04.H20 (aq) + H20 (1) ercentage yield of crude product (remember to include the 0.025 g of compound you take for determining the absorption spectrum). actual yieldx100 ed theoretical yield

Answer 1

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Answer CuSO: 5H, 0 (s) + 4N Hylan) Teaca Ha),]$ 0,: Holes) + Hoce) 249.72 g/mol + 124 245.75 Istoichiometric Xati weights in gms) (5.011 gm of Cusoy .5H2O) x (umul. Cuso il cond) x 249.42 g Cu Sou SH₂O) A (amol. [Cu(NH 3) u] Sou. Ho V245.759 of [Cu(NH3), Southe) - 1 mol. lusov.rH₂O 1 I mo of [CHUNH3) Soy H2o - 4.920 gm of (cu (NH) u J Soy. H₂O the outical or experimental yield is 4.920gm. limiting reactant limiting reactant can be calculated from the reaction to see no. of moles of reactant which reactant has dels no. of moles is called as limiting reactant Cusou. 5 H2O is here a limiting reaitant because it has one mole and NH₂ has 4 moles.

2 % yield = autual yield theoretical yield 1.831 X 100 4.920 0.37 21 X 100 37.210%