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Question 1 of 3

What is the percentage yield of Ethyne in the following reaction:

CaC2 + 2H2O ---------» C2H2 + Ca(OH)2
if 10 grams of Calcium Carbide are used and 3 litres of Ethyne are produced - measured at 0°C and 760 mm Hg? Answer to 1 decimal place.
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%
The percentage yield of Ethyne is 85.8 %.
  1. Find the theoretical yield of Ethyne.
    1. Find the moles of Calcium Carbide used, where Moles = Mass/RMM         RMM = 64.1 Moles of Calcium Carbide used = 10/64.1 = 0.15600 moles
    2. Find the theoretical volume of Ethyne that could be produced. Use the equation:
      1 mole of Calcium Carbide could produce 1 mole of Ethyne. Thus:
      0.15600 moles of Calcium Carbide could produce 0.15600 moles of Calcium Carbide
      At 0°C and 760 mm Hg, 1 mole of Ethyne has a volume of 22.4 litres. Thus:
      0.15600 moles of Ethyne has a volume of 0.15600 x 22.4 litres = 3.4945 litres
  2. Find the percentage yield of Ethyne.
    Percentage Yield = Actual Yield/Theoretical Yield x 100 %
    Percentage Yield of Ethyne = 3 / 3.4945 x 100 % = 85.84 % = 85.8 % to 1 dec pl

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Question 2 of 3

0.76 grams of Potassium Sulphate are produced using 70 ml of 0.2000 M Potassium Hydroxide and 46 ml of 0.1000 M Sulphuric Acid in the following reaction:

2KOH + H2SO4 ---------» K2SO4 + 2H20
What is the percentage yield of Potassium Sulphate? Answer to 1 decimal place.
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%
The percentage yield of Potassium Sulphate is 94.8 %.
  1. Find the theoretical yield of Potassium Sulphate.
    1. Find which of the reactants is limiting.
      Molarity = Moles/Volume (in litres): Moles = Molarity x Volume (in litres)
      Moles of Potassium Hydroxide   = 0.2000 x 0.070 = 0.014 moles
      Moles of Sulphuric Acid             = 0.1000 x 0.046 = 0.0046 moles
      From the balanced equation you can see that:
      2 moles of Potassium Hydroxide could produce 1 mole of Potassium Sulphate. Thus:
      0.014 moles of Potassium Hydroxide could produce 0.014/2 = 0.007 moles of Sulphate
      1 mole of Sulphuric Acid could produce 1 mole of Potassium Sulphate. Thus:
      0.0046 moles of Sulphuric Acid could produce 0.0046 moles of the Sulphate.
    2. Theoretical Mass of Sulphuric Acid = Moles x RMM = 0.0046 x 174.2 g = 0.80132 g
  2. Find the percentage yield of Potassium Sulphate.
    Percentage Yield = Actual Yield/Theoretical Yield x 100 = 0.76/0.80132 x 100 = 94.8 %

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Question 3 of 3

In the reaction:
C2H5COOH + C3H7OH ---------» «--------- C2H5COOC3H7 + H2O

111 grams of Propanoic Acid reacts with 105 grams of Propanol to produce 113 grams of Propyl Propanoate. What is the percentage yield of Propyl Propanoate? Answer to 1 decimal place.

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%
The percentage yield of Propyl Propanoate is 64.9 %.
  1. Find the theoretical yield of Propyl Propanoate.
    1. Find which of Propanoic Acid and Propanol is the limiting reactant.
      Moles of Propanoic Acid   = Mass/RMM = 111/74.0 = 1.5 moles       RMM = 74.0
      Moles of Propanol             = Mass/RMM = 105/60.0 = 1.75 moles     RMM = 60.0
      From the balanced equation, you can see that:
      1 mole of Propanoic Acid could produce 1 mole of Propyl Propanoate. Thus:
      1.5 moles of Propanoic Acid could produce 1.5 moles of Propyl Propanoate.
      1 mole of Propanol could produce 1 mole of Propyl Propanoate. Thus:
      1.75 moles of Propanol could produce 1.75 moles of Propyl Propanoate.
    2. Theoretical Mass of Propyl Propanoate = Moles x RMM = 1.5 x 116.0 = 174 g
  2. Find the percentage yield of Propyl Propanoate.
    Percentage Yield = Actual Yield/Theoretical Yield x 100 %
    Percentage Yield = 113/174 x 100 % = 64.94 % = 64.9 % to 1 dec pl

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