limiting reactant and theoretical yield - … · i. limiting reactant one reactant will be used up...
TRANSCRIPT
LIMITING REACTANT AND THEORETICAL
YIELD
Lesson 5.4
I. Limiting Reactant
� What if the reactants are not present in the ratio you need?
I. Limiting Reactant
�Normally, in nature or the lab, reacting chemicals are not present in perfect whole number ratios.
�SHOCKING….but true.
I. Limiting Reactant
� One reactant will be used up first.� Limiting reactant – the reactant that
exists in the smallest proportional amount, according to the mole ratio.
� The other reactant will have some left over.
I. Limiting Reactant
� Reaction of hydrochloric acid (HCl) and Zinc
II. Theoretical Yield� Predicting how much product will result from a
quantity of reactants.� Three Steps:� 1) Calculate the yield (moles or grams of
product) as if the 1st reactant is limiting.� 2) Calculate the yield as if the 2nd reactant is
limiting.� 3) The smaller of the two numbers is the
theoretical yield.� * The reactant that gives the smallest number
is the limiting reactant.
II. Theoretical Yield
� Skateboard factory example:� You inherit a skateboard factory. The factory
has some items still in stock.� You have:
¡167 wheels¡93 trucks¡71 decks
� How many skateboards can you make?
II. Theoretical Yield
� 4 wheels + 2 trucks + 1 deck à 1 skateboard
� Theoretical yield is 41 skateboards� The limiting reactant is the wheels.
Skatebds 41 wheels4Skatebd 1 wheels167 =×
Skatebds 46 trucks2
Skatebd 1 trucks93 =×
Skatebds 71board 1
Skatebd 1 boards 71 =×
II. Theoretical Yield
� Chem. Example:� 2 Ag(s) + I2(s) à 2 AgI(s)� If you start with 1.00 g of each reactant, what is
the theoretical yield (in grams)?� Reminder: You MUST convert to moles!
II. Theoretical Yield
�Step 1:� 1.00 g Ag à ? g AgI
�Possible yield = 2.18 g AgI
AgIgAgImolAgIg
AgmolAgImol
AggAgmolgAg 18.2
1 235
2 2
108 100.1 =×××
II. Theoretical Yield
�Step 2:� 1.00 g I2 à ? g AgI
�Possible yield = 1.85 g AgI
AgIgAgImolAgIg
ImolAgImol
IgImolgI 85.1
1 235
1 2
542 100.1
22
22 =×××
II. Theoretical Yield
�Step 3:�TY = 1.85 g AgI�LR is iodine�There is extra silver
�The actual yield would likely be less due to unwanted “side” rxns and product loss.
Practice
� Mole ratio = 1:1:1:1:1� Step 1: 5.24 g CO2
� Step 2: 7.33 g CO2
� TY = 5.24 g CO2
� LR = NaHCO3
IV. Percent Yield
�% yield = Actual yieldTheoretical Yield
� If 4.57 g of CO2 was formed in an actual trial, what is the percent yield?