IB DP Chemistry: SL复习笔记1.2.2 Reaction Yields

Reaction Yields

Percentage yield

  • In a lot of reactions, not all reactants react to form products which can be due to several factors:
    • Other reactions take place simultaneously
    • The reaction does not go to completion
    • Products are lost during separation and purification
  • The percentage yield shows how much of a particular product you get from the reactants compared to the maximum theoretical amount that you can get:

1.2.2-Percentage-yield-formula

  • The actual yield is the number of moles or mass of product obtained experimentally
  • The theoretical yield is the number of moles or mass obtained by a reacting mass calculation

Worked Example

In an experiment to displace copper from copper(II)sulfate, 6.5 g of zinc was added to an excess of copper(II)sulfate solution.The resulting copper was filtered off, washed and dried.The mass of copper obtained was 4.8 g.Calculate the percentage yield of copper.

Answer:

Step 1: The symbol equation is:

Zn (s)+CuSO4 (aq)→ZnSO4 (aq)+ Cu (s)

Step 2: Calculate the amount of zinc reacted in moles

1.2.2-Worked-Example-Moles

Step 3: Calculate the maximum amount of copper that could be formed from the molar ratio:

Since the ratio of Zn(s) to Cu(s) is 1:1 a maximum of 0.10 moles can be produced

Step 4: Calculate the maximum mass of copper that could be formed (theoretical yield)

mass=molxM=0.10molx63.55g mol-1=6.4 g (2 sig figs)

Step 5: Calculate the percentage yield of copper

1.2.2-Worked-Example-Percentage-yield

 

 

 

 

 

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