Question

EXPERIMENT 8: Synthesis of Aquapentaammine-Cobalt (II) Nitrate, [Co(NH)sH2O1(NO)s Pre-Lab Questions: The complex salt [Cu(NH3)4][SO,1- H,O can be prepared by direct combination of its constituents: Cu +4 NH,+SO+ H2O [Cu(NH)JSO. H;O In an aqueous reaction solution contained 1.0 g CuCl, 2H2O, 1.0 mL 14 M NHs, and 1.0 g NazSO, what would be the maximum theoretical vield (in grams) of product? To what percent yield would 0.42 g of product correspond? Your yield of crude product is 1.8 g and you need 20 mL of 1 M NH3 (aq) per gram of complex for the recrystallization step. How many mL of 1 M NH3 (aq) will you need? How much 6 M NH (aq) will you need to dilute to make this amount of 1 M solution? 2. 3. For the yield of product given in Prelab Problem 2, what amount of 16 M nitric acid would need to be added to the solution to neutralize the added NH3 (aq)?

Answer 1

1. Complex salt : [Cu(NH₃)₄]SO₄.H₂O

-we have : 1.0 g CuCl₂.2H₂O : moles = 1.0 g/ 170.48g/mol = 5.6*10^-3 mole

So, moles of Cu²⁺ = 5.6*10^-3 mole

- 1 ml 14 M NH₃ : moles of NH₃ = (1ml/ 1000ml/L)*14 mol/L= 0.014 moles

- 1 g Na₂SO₄ :  moles = 1.0 g/ 142 g/mol = 7.04*10^-3 mole

( 1 mol of Na₂SO₄ : 1 moles of SO₄^2- )

So, moles of SO₄^2- = 7.04*10^-3 mole

- water is in abundance .

Theoretical yield is determined by limiting reagent :

Required mole ratio = Cu²⁺ :NH₃ :SO₄^2-  = 1:4:1

we have , Cu²⁺ :NH₃ :SO₄^2-  = 1: 2.5: 1.26

So limiting reagent is ammonia :

theoretical yield = 0.014 mole/ 4 = 3.5*10^-3 mole

Max. theoretical yield = 3.5*10^-3 mole * 245.79 g/mol = 0.86 g

% yield for 0.42 g = 0.42g /0.86g *100 = 48.84 %

2. Crude product = 1.8 g

needed 20 ml 1 M NH₃ (aq) /g of product for re-crystallization

So, for 1.8 g we need = 36 ml of 1 M NH₃ (aq).

If we had  6 M NH₃ (aq) : to get 1 M NH₃ (aq) we need to dilute 6 times. (dilution factor = 6).

3.

we have, NH₃ (aq) + HNO₃ (aq) ==>  NH₄NO₃ (aq)

we added 36 ml of 1 M NH₃ (aq)​​​​​​​ or 0.036 moles

we will need 0.036 moles of  HNO₃ (aq) for neutralization :

16 M HNO₃ (aq) , : (0.036 mole / 16 mole/L )*1000 ml/L = 2.25 ml.