Preparation of Ammine Complexes of Cobalt(III) Part A Pentammine chlorocobalt(II

Question

Preparation of Ammine Complexes of Cobalt(III)

Part A Pentammine chlorocobalt(III) chloride-

[Co(H2O)6]2+ +5NH3 = [Co(NH3)5(H2O)]2+ + 5H2O

2[Co(NH3)5(H2O)]2+ + H2O2 --> 2[Co(NH3)5(OH)]2+ +2H2O

[Co(NH3)5(OH)]2+ +3Cl- --> [Co(NH3)5Cl]Cl2(s) + 2OH-

Preparation of pentammine chlorocobalt(III) Chloride-[Co(NH3)5CI]Cl2

Moles of Cobalt(III) Chloride Hexahydrate

Please show how to do calculations.

Weight of Cobalt(II) Chloride Hexahydrate 5.190g

Moles of Cobalt(III) Chloride Hexahydrate

? Weight of Recovered Product, Filter Paper and Watch Glass 54.643g Weight of Filter Paper and Watch Glass 51.841g Experimental Yield of Product ? Theoretical Yield of Product ? Percent Yield of Product ?

Explanation / Answer

Mol. wt of Cobalt(II) Chloride Hexahydrate = 237.93 g/mol

Weight of Cobalt(II) Chloride Hexahydrate 5.190g

A. hence, moles of Cobalt(II) Chloride Hexahydrate = 5.190/237.93 = 0.0218 mol

B. Weight of Recovered Product, Filter Paper and Watch Glass = 54.643g

Weight of Filter Paper and Watch Glass= 51.841g

Hence, weight of recovered pdt i.e. pentammine chlorocobalt(III) Chloride =(54.643-51.841) g = 2.802 g = experimental yield.

C. Theoretically 1 mol of Cobalt(II) Chloride Hexahydrate should form 1 mol of pentammine chlorocobalt(III) Chloride

i.e. 0.0218 mol of Cobalt(II) Chloride Hexahydrate should form 0.0218 mol = 0.0218 mol * 250.4 g/mol [mol. wt of pentammine chlorocobalt(III) Chloride = 250.4 g/mol] = 5.459 g of pentammine chlorocobalt(III) Chloride

Hence, theoretical yield of product = 5.459 g

D. Percent yield of product = 2.802 g/5.459 g * 100% = 51.32%

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