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Question 1 of 4

What is the theoretical yield of Ethane in litres (measured at 0°C and 760 mm Hg) in the following reaction if 12.1 grams of Ethyl Bromide are used? Answer to 3 decimal places.

C2H5Br + Mg ---------» C2H5MgBr
      C2H5MgBr + H2O ---------» C2H6 + MgBr(OH)
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litres
The theoretical yield of Ethane is 2.489 litres.
  1. Find the moles of Ethyl Bromide used.
    Moles of Ethyl Bromide = Mass / RMM                                      RMM = 108.9
    Moles of Ethyl Bromide = 12.1 /108.9 = 1/9 moles
  2. Find the number of moles of Ethane that can be produced:
    Looking at both equations:
    1 mole of Ethyl Bromide produces 1 mole of Ethyl Magnesium Bromide
      and 1 mole of Ethyl Magnesium Bromide produces 1 mole of Ethane.
    Thus 1 mole of Ethyl Bromide leads to the production of 1 mole of Ethane.
    Thus 1/9 moles of Ethyl Bromide leads to the production of 1/9 moles of Ethane.
  3. Find the volume of Ethane that can be produced, measured at 0°C and 760 mm Hg.
    1 moles of Ethane has a volume of 22.4 litres. Thus:
    1/9 moles of Ethane has a volume of 1/9 x 22.4 litres = 2.4888 litres
    1/9 moles of Ethane has a volume of 2.489 litres to 2 dec pl

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Question 2 of 4

You need to produce 24.58 grams of Propyl Bromide. The percentage yield of this reaction using the equipment available is 86.0 %. What is the minimum amount in grams of Propanol that you need? Answer to 2 decimal places.

C3H7OH + HBr ---------» C3H7Br + H2O
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grams
The minimum amount of Propanol that you need is 13.95 grams.
  1. The percentage yield is 86.0 %. Find the theoretical yield of Propyl Bromide needed.
    Percentage Yield = Actual Yield / Theoretical Yield x 100% Thus:
    Theoretical Yield = Actual Yield / Percentage Yield x 100 grams
    Theoretical Yield of Propyl Bromide = 24.58 / 86.0 x 100 grams = 28.581 grams
  2. Find the number of moles of Propyl Bromide needed. RMM = 122.9
    Moles of Propyl Bromide needed = Mass / RMM = 28.581/122.9 = 0.23255 moles
  3. Find the number of moles of Propanol needed.
    From the balanced equation you can see that:
    1 mole of Propyl Bromide is produced from 1 mole of Propanol
    Thus 0.23255 moles of Propyl Bromide are produced from 0.23255 moles of Propanol.
  4. Find the mass of Propanol needed. RMM = 60.0
    Mass of Propanol needed = Moles x RMM
    Mass of Propanol needed = 0.23255 x 60.0 g = 13.953 g = 13.95 g to 2 dec pl

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Question 3 of 4

What is the theoretical yield in grams of 1-2 dibromopropane in the following reactions:

conc H2SO4110 - 115°CCH3CHOHCH3 --------------» CH3CHCH2
             C3H6 + Br2 --------------» CH3CHBrCH2Br
if 15 grams of propan-2-ol are used. Answer to 2 decimal places.
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grams
The theoretical yield of 1-2 dibromopropane is 50.45 grams.
  1. Find the moles of Propan-2-ol used.
    Moles = Mass / RMM                                                      RMM = 60.0
    Moles of Propan-2-ol = 15/60.0 = 0.25 moles
  2. Find the moles of 1-2 dibromopropane that can be produced.
    From the equations you can see that:
    1 mole of Propan-2-ol produces 1 mole of Propene.
    1 mole of Propene produces 1 mole of 1-2 dibromopropane
    Thus 1 mole of Propan-2-ol leads to the production of 1 mole of 1-2 dibromopropane. And
    0.25 moles of Propan-2-ol leads to the production of 0.25 moles of 1-2 dibromopropane.
    Thus 0.25 moles of 1-2 dibromopropane can be produced.
  3. Find the mass of 1-2 dibromopropane that can be produced.
    Mass = Moles x RMM                                                      RMM = 201.8
    Mass of 1- dibromopropane = 0.25 x 201.8 g = 50.450 g = 50.45 g to 2 dec pl

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Question 4 of 4

What is the percentage yield of Ethyl Iodide from the following reactions if 31.0 grams of Phosphorus are used and 320.0 grams of Ethyl Iodide are produced? No other reactants are limiting. Answer to 1 decimal place.

2P + 3I2 ---------» 2PI3
        3C2H5OH + PI3 ---------» 3C2H5I + H3PO3
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%
The percentage yield of Ethyl Iodide is 68.4 %.
  1. Find the theoretical yield of Ethyl Iodide.
    1. Find the moles of Phosphorus used.
      Moles of Phosphorus = Mass / RMM = 31.0 / 31.0 = 1 mole                  RMM = 31.0
    2. Find the moles of Ethyl Iodide that can be produced from 1 mole of Phosphorus.
      From the balanced equations you can see that:
      2 moles of Phosphorus can produce 2 moles of Phosphorus Iodide. Thus
      1 mole of Phosphorus can produce 1 mole of Phosphorus Iodide.
      and 1 mole of Phosphorus Iodide can produce 3 moles of Ethyl Iodide. Thus:
      1 mole of Phosphorus can produce 3 moles of Ethyl Iodide.
    3. Therefore mass of Ethyl Iodide that can be produced is:
      Mass of Ethyl Iodide = Moles x RMM = 3 x 155.9 g = 467.7 g
  2. Find the percentage yield of Ethyl Iodide (%age Yield = Actual Yield/Theoretical Yield x 100)
    Percentage Yield = 320.0 / 467.7 x 100 % = 68.41 % = 68.4 % to 1 dec pl

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