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?
Answer:
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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