Question

Consider:

2CoCl2(aq) + 8NH3(aq) + H2O2(aq) + 2NH4Cl(aq) → 2[Co(NH3)5Cl]Cl2(s) + 2H2O(l)

In a 50 mL Erlenmeyer flask were combined 1.075 g of cobalt(II) chloride hexahydrate, 1.2 mL of

15 M aqueous solution of ammonia, 0.505 g of ammonium chloride, 1.0 mL of 30.0 % H2O2 (density = 1.10 g/mL), and 5.0 mL of 12 M hydrochloric acid. Reaction described by the equation above occurred, and isolated was 0.450 g of the cobalt-containing product shown. What was the percent yield of product?

Answer 1

Ans. Part A: Calculate moles of all the reactants

# Moles of CoCl₂ = Mass / Molar mass = 1.075 g / (129.8386 g/mol) = 0.00828 mol

# Moles of NH₃ = Molarity x Volume in liters = 15 M x 0.0012 L = 0.0180 mol

# Moles of NH₄Cl = 0.505 g / (53.4912 g/mol) = 0.009441 mol

# Mass of H₂O₂ solution = Volume x density = 1.0 mL x (1.10 g/ mL) = 1.10 g

Mass of H₂O₂ = 30% of mass of H₂O₂ solution = 0.30 x 1.10 g = 0.33 g

Moles of H₂O₂ = 0.33 g / (34.01468 g/mol) = 0.009702 mol

# Note that HCl does not take part in reaction because it’s not shown in the balanced reaction.

#Part B: Determine the limiting reactant:

# Theoretical molar ration of reactants according to the stoichiometry of balanced reaction is-

            CoCl₂ : NH₃ : H₂O₂ : NH₄Cl = 2 : 8 : 1 : 2

# Experimental molar ration of reactants is-

            CoCl₂ : NH₃ : H₂O₂ : NH₄Cl = 0.00828 : 0.0180 : 0.009702 : 0.09441

= 2 : 4.34 : 2.28 : 2.34

# Compare the theoretical and experimental molar ratio of reactants, the experimental moles of NH₃ is less than its theoretical value of 8 mol, whereas the experimental moles of all other reactants is equal to or greater than their respective theoretical values.

So,

            NH₃ is the limiting reactant.

# Calculate theoretical yield.

The formation of product follows the stoichiometry of limiting reactant.

According to the stoichiometry of balanced reaction, 8 mol NH₃ produces 2 mol [Co(NH₃)₅Cl]Cl₂.

So,

            Theoretical moles of [Co(NH₃)₅Cl]Cl₂ formed = (2 / 8) x Moles of NH₃

                                                                                    = (2 / 8) x 0.0180 mol

                                                                                    = 0.0045 mol

Theoretical mass of [Co(NH₃)₅Cl]Cl₂ formed = Theoretical moles x Molar mass

                                                                                    = 0.0045 mol x (250.4441 g/ mol)

                                                                                    = 1.127 g

# Calculate % yield

            % yield = (Actual yield / Theoretical yield) x 100

                        = (0.450 g / 1.127 g) x 100

                        = 39.93 %