# Consider: 2CoCl2(aq) + 8NH3(aq

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 Mhydrochloric acid. Reaction described by the equation aboveoccurred, and isolated was 0.450 g of the cobalt-containing productshown. What was the percent yield of product?

Ans. Part A: Calculate moles of all thereactants

# 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 x0.0012 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 ofH2O2 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 notshown in the balanced reaction.

#Part B: Determine the limiting reactant:

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

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

# Experimental molar ration of reactantsis-

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 ofreactants, the experimental moles of NH3 is less thanits theoretical value of 8 mol, whereas the experimental moles ofall other reactants is equal to or greater than their respectivetheoretical values.

So,

NH3 is the limiting reactant.

# Calculate theoretical yield.

The formation of product follows the stoichiometry of limitingreactant.

According to the stoichiometry of balanced reaction, 8 molNH3 produces 2 mol[Co(NH3)5Cl]Cl2.

So,

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

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