Limiting reactant reactions

1
Limiting reactant reactions

• Reaction where you are given two amounts to start
  with
• Usually both reactants
• Asks for how much product can be produced

2
The reactants

• Limiting reactant
• The reactant that is completely consumed in a
  reaction
• It stops the reaction from producing any more
  products
• Excess reactant
• The reactant that is not completely consumed
• There is left over reactant

3
Real life example

• Making a sandwich

4
How to solve

• Example
• H2 O2 ? H2O
• If you have 5 moles of hydrogen and 5 moles of
  oxygen, how much water can be produced?

5
Step 1

• Balance the equation
• __H2 __O2 ? __H2O

6
Step 2

• Convert the amount of each reactant given to
  moles (if needed)
• How?
• grams given 1 mol molecule
• --------------------- ---------------------
  moles
• 1 g of molecule
• In our example, we are already in moles, so we do
  not have to do this step
• 5 moles hydrogen
• 5 moles oxygen

7
Step 3

• Determine the limiting reactant
• Divide the amount of moles of each reactant by
  their corresponding coefficient in the balanced
  equation
• This will give you a ratio of the amount of
  reactant you have
• The smaller ratio is the limiting reactant
• In our example
• 5 moles of hydrogen / 2 2.5
• 5 moles of oxygen / 1 5

8
Step 4

• Now that you know your limiting reactant, we use
  the amount of this reactant given in the problem
  to solve for the amount of product
• YOU ALWAYS USE THE AMOUNT GIVEN IN THE PROBLEM,
  NOT THE ANSWER TO STEP 3.
• The answer to step three is only used to tell
  which one is the limiting reactant.
• 5 mols H2 2 mols H2O
• -------------- x --------------- 5 mols
  water
• 1 2 mols H2

9
What if I want to do two separate Stoichiometry
problems?

• If you do, then that is ok. Just remember that
  your amount of product that can be produced has
  to be the smaller answer
• 5 mols H2 2 mols H2O
• -------------- x --------------- 5 mols
  water
• 1 2 mols H2
• 5 mols O2 2 mols H2O
• -------------- x --------------- 10 mols
  water
• 1 1 mols O2

10
Example

• Na Cl2 ? NaCl
• Suppose 5.40 moles of Na react with 2.90 moles of
  Cl2. How much NaCl (in moles) will be produced?
• 5.40 moles Na / 2 2.7
• 2.90 moles Cl2 / 1 2.9
• 5.4 mols Na 2 mols NaCl
• -------------- x --------------- 5.4 mols
  NaCl
• 1 2 mols Na

11

• Na Cl2 ? NaCl
• Suppose 75.40 g of Na react with 26.90 g of Cl2.
  How much NaCl (in moles) will be produced?

12
Practice with mass

• 1
• 28 grams N2 1 mol N2
• --------------------- ---------------------
  1mol N2
• 1 28 g N2
• 25 grams H2 1 mol H2
• --------------------- ---------------------
  12.5 moles H2
• 1 2 g H2

13
Bell work

• Na Cl2 ? NaCl
• If 50 grams of sodium and 78 grams of chlorine
  react, how much (in grams) sodium chloride is
  produced?
• Be sure to identify the limiting reactant and
  solve for the amount of NaCl that can be produced.

14
Answer

• 2 Na 1 Cl2 ? 2 NaCl
• 50 grams Na 1 mol Na
• --------------------- ---------------------
  2.174 mol Na
• 1 23 g Na
• 78 grams Cl2 1 mol Cl2
• --------------------- ---------------------
  1.114 moles H2
• 1 70 g Cl2

15
Answer continued

• 2.17 mol Na / 2 1.085 (limiting)
• 1.11 mol Cl2 / 1 1.11
• 50 g Na 1 mol Na 2 mol NaCl 58 g NaCl
• ------------- x ----------- x -------------
  x---------------
• 1 23 g Na 2 mol Na 1
  mol NaCl
• 126 g NaCl gt 100 g NaCl

16
Practice

• A sample of potassium and bromine react to form a
  product. If 37.1 grams of potassium is used and
  42.5 grams of bromine, then how much product (in
  grams) was formed.
• Write the balanced equation.
• What type of reaction is this?
• Determine the limiting reactant.
• Solve for the amount of product formed.

17

• 2K Br2 ? 2KBr
• 37.1 grams K 1 mol K
• --------------------- ---------------------
  0.95 mol K
• 1 39 g K
• 42.5 grams Br2 1 mol Br2
• --------------------- ---------------------
  0.266 moles Br2
• 1 160 g Br2

18
2K Br2 ? 2KBr

• 0.95 mol K / 2 0.475
• 0.266 moles Br2 / 1 0.266 limiting
• 42.5 grams Br2 1 mol Br2 2 mol KBr
  119 g KBr
• ------------------- x --------------- x
  ------------------ x------------------
• 1 160 g Br2 1
  mol Br2 1 mol KBr
• 63.2 g KBr

19
When the reaction is done

• How much of each reactant is left?
• Limiting
• None is left
• It is completely used up
• Excess
• You have to determine how much is left
• How?
• Use the answer from the Stoichiometry problem to
  solve for the amount of excess reactant that was
  used
• Subtract the amount used from the amount started
  with

20
Lets see an example to do this

• Cu S ? Cu2S
• How much product (in moles) can be formed if 74.0
  g Cu reacts with 23.0 g S?
• 74 grams Cu 1 mol Cu
• --------------------- ---------------------
  1.156 moles Cu
• 1 64 g Cu
• 23 grams S 1 mol S
• --------------------- ---------------------
  0.718 moles S
• 1 32 g S

/ 2 0.578
/ 1 0.718
21

• 74 grams Cu 1 mols Cu 1 mol Cu2S
• --------------- x -------------- x
  --------------- 0.58
• 1 64 g Cu 2
  mols Cu mols
• 
  Cu2S
• Now that we have an answer for our product, lets
  use it to determine how much excess reactant was
  used
• 0.58 mols Cu2S 1 mols S 32 g S
• -------------------- x -------------- x
  ----------------
• 1 1 mol Cu2S 1 mol S
• 18.56 g S this is how much was used, now we
  need to subtract it from the amount of excess
  we started
• with

22

• 23.0 g S
• 18.56 g S
• -----------------------
• 4.44 g S is left over
• after the reaction is over

23
Lets put it all together - Last one

• Hydrochloric acid reacts with Aluminum hydroxide
  to form Aluminum chloride and water.
• Write the complete chemical reaction
• Classify the type of reaction
• If 40.0 g of Hydrochloric acid and 15.0 g
  Aluminum hydroxide react together, which reactant
  is the limiting reactant?
• Use the limiting reactant to determine how many
  grams of water will be produced.
• Tell me how much excess reactant is left over
  once the reaction has completed.

24
Last topic in unit 4 percent yield

• So far, we have used Stoichiometry to predict how
  much product or reactant was needed in a reaction
• The answers to the Stoichiometry problems have a
  special name
• Theoretical value or yield the amount the
  Stoichiometry problem predicted the reaction
  should produce
• We do not always get the theoretical amount of
  product we expect in an experiment
• That is why we call the amount produced in the
  lab the
• Experimental value or yield the amount you
  experimentally determine in the lab

25
Percent Yield

• We need a way to compare the theoretical and
  experimental values
• This is why we will use percent yield
• Tells us how much of the product you actually
  produced in the lab
• Tells us how right you were during the
  experiment
• Formula Experimental value
• ------------------------- x
  100
• Theoretical value

26
Practice

• Pb(NO3)2 ? Pb NO2 O2
• Balance the equation
• What type of reaction is this?
• If you have 50 grams of lead (II) nitrate, how
  many grams of nitrogen dioxide can be produced?
  This is your theoretical value
• If during a lab experiment, you were only able to
  produce 7 grams of NO2, what is your percent
  yield?

27
Bell Work

• A student burned 15.5 g oxygen with 165 g of iron
  to make iron (III) oxide.
• Write a balanced equation for this reaction.
• Classify the type of reaction.
• Use the amounts given to determine the limiting
  reactant.
• Use the limiting reactant to solve for grams of
  the iron (III) oxide. (Determine the theoretical
  yield for the reaction.)
• If you did this experiment in the lab and you
  were able to form only 10 grams of iron (III)
  oxide, what would your yield be for this
  reaction?

28
Review from yesterday

• Cu S ? Cu2S
• How much product (in moles) can be formed if
  100.0 g Cu reacts with 50.0 g S?
• Determine the limiting reactant, use that to
  solve for moles of the product, then use this
  answer to solve for the excess reactant was used
  and finally, how much excess was left over at the
  end of the reaction.
• 100 grams Cu 1 mol Cu
• --------------------- ---------------------
  1.5625 moles Cu
• 1 64 g Cu
• 50 grams S 1 mol S
• --------------------- ---------------------
  1.5625 moles S
• 1 32 g S

/ 2 0.78125
/ 1 1.5625
29

• 100.0 grams Cu 1 mols Cu 1 mol Cu2S
• --------------- x -------------- x
  --------------- 0.7813
• 1 64 g Cu 2
  mols Cu mols
• 
  Cu2S

0.7813 mols Cu2S 1 mols S 32 g
S -------------------- x -------------- x
---------------- 1 1 mol Cu2S
1 mol S 25.00 g S ? were used 50.0
g (given) 25.00 g (used) 25.0 g S left over
30
Practice

• Balance and tell me what type of reaction is
  occurring.
• CH4 O2 ? CO2 H2
• Ba H2O ? Ba(OH)2 H2
• C H2 ? C2H2