Synthesis of carbonatotetraaminecobalt(III) nitrate [Co(NH 3 ) 4 CO 2 ]NO 3 Diss

Question

Synthesis of carbonatotetraaminecobalt(III) nitrate [Co(NH3)4CO2]NO3

Dissolve 20 g (0.21 moles) of ammonium carbonate in 60 mL of water, and to this add 60 mL of concentrated aqueous ammonia. Prepare a second solution by dissolving 15 g (0.052 moles) of cobalt(II) nitrate in 30 mL of water. With stirring, add the first solution to the second solution, and then slowly add 8 mL of 30% hydrogen peroxide.

Place the resulting solution on a hot plate, and reduce the volume to between 90 and 100 mL (this should be performed in a fume hood). During this stage, add in small portions 5 g (0.05 moles) of ammonium carbonate. While solution is still hot, perform vacuum filtration and cool the filtrate in an ice–water bath. Red crystals will precipitate out, and are collected by vacuum filtration. Wash the crystals with ice cold water and then ethanol.

Record the mass of crystals collected and calculate the percent yield. The crystals are carbonatotetraaminecobalt(III) nitrate.

How can I calculate the percent yield if i got 7.306g of final product???

Explanation / Answer

Ans. Balanced reaction:

10 Co(NO3)2 + 26 NH3 + 10 (NH4)2CO3 + H2O2 = 10 [Co(NH3)4CO3]NO3 + 8 NH4NO3OH

Given reagents:

            (NH4)2CO3 = 0.21 moles

            Co(NO3)2 = 0.052 moles

            NH3 = excess

Note the stoichiometry of balanced reaction, ammonium carbonate and cobalt nitrate are consumed in equimolar ration (10 moles each). However, under given experimental conations, moles of cobalt nitrate is lesser than moles of ammonium carbonate.

Therefore, cobalt nitrate, Co(NO3)2 is the limiting reagent. Also, the product formation follows stoichiometry of limiting reagent.

Note the stoichiometry of balanced reaction, 10 mol Co(NO3)2 produces 10 mol [Co(NH3)4CO3]NO3. That is, 1 mol limiting reactant produces 1 mol desired product.

Therefore,

Theoretical moles of [Co(NH3)4CO3]NO3 formed

= Theoretical moles of Co(NO3)2 consumed = 0.052 mol

Theoretical yield of [Co(NH3)4CO3]NO3 = Theoretical moles produced x Molar mass

                                                            = 0.052 mol x (249.06958 g/mol)

                                                            = 12.952 g

% yield of [Co(NH3)4CO3]NO3 = (Observed yield / theoretical yield) x 100

                                                = (7.306 g / 12.952 g) x 100

                                                = 56.41 %

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