Limiting Reactants

Chapter 13

A real-life example• Cookies anyone?• Given the following recipe for chocolate

chip cookies…

• What if we only have 1 egg? How many cookies could we make? What would we need to do? What would we have left over?

Demo: Bolts & Washers

2Mg + O2 2MgO

•1 bolt = 1 mole of Mg

•2 washers = 1 mole of O2

•What would 1 mole MgO look like?

•Let’s “make” our reaction with bolts and washers…

Demo: Bolts & Washers

+=

2Mg + O2 = 2MgO

Demo: Bolts & Washers

• When does that “reaction” stop?

• What did you run out of?– Liming Reactant

• What is left over?– Excess Reactant

Definitions• Reactant

– Is a substance that takes part in a chemical reaction to make a product

• Limiting Reactant– The reactant that is completely used up

in the reaction and limits the amount of product

• Excess Reactant– The reactant that is in excess, not all

used up.

Why do we care? (Don’t write…let’s discuss)

• Scientific Reasons– Determine how much reactant you

need – Determine how much product will be

produced

• Economic (Show me the money!)– Keep production cost’s minimal– Maximize output in production

Let’s Review…

Steps in solving stoichiometry problems…(how do we find mass or volume of a product given mass or volume of a reactant?)

1.Write and Balance Equation

2.Identify known and unknown

3.Convert mass (or volume) to moles

4.Use mole ratio to convert from moles of known to moles of unknown

5.Convert moles of unknown to mass (or volume) of unknown

Steps in Solving LR problems!

When two amounts of reactants are given in a problem, we need to identify the limiting reactant to solve for the amount of product possible!

1.1. Start with a Balanced EquationStart with a Balanced Equation

2.2. Convert grams of each reactant to Convert grams of each reactant to moles (this is how much you have of moles (this is how much you have of each)each)

3.3. Use mole ratio to convert moles of one Use mole ratio to convert moles of one react to moles of the other and react to moles of the other and compare to the original number of compare to the original number of moles obtained in step two (what you moles obtained in step two (what you need)need)

4.4. Decide which is the limiting reactantDecide which is the limiting reactant

Another Method

1. Always begin with a BALANCED EQUATION!!

2. Convert the amount (grams/volume) of each reactant to moles of reactants.

3. Convert the number of moles of reactant to moles of product (doesn’t matter which one as long as you use the same product for both reactants)

4. Which ever reactant produces the least product is the limiting reactant.

Example 1• The reaction begins with 2.51g of HF The reaction begins with 2.51g of HF

and 4.56g of SiOand 4.56g of SiO22. .

HF + SiOHF + SiO22 SiF SiF44 + H + H22OO• Balance EquationBalance Equation• Convert g Convert g mols of each (What you have) mols of each (What you have)• Pick one & convert mols Pick one & convert mols moles of other moles of other

reactant. This is what you need…Do you have reactant. This is what you need…Do you have enough?enough?

• If If ““nono””, then that is your Limiting Reactant… if , then that is your Limiting Reactant… if yes than the other is your LRyes than the other is your LR

2. Calculate moles of each reactant (what you have)

2.51 g HF 1 mol HF = 0.1255 mol HF20.0 g HF

4.56 g SiO2 1 mol SiO2 = 0.059 mol SiO2

76.05 g SiO2

Example 11. Balance the Equation1. Balance the Equation

4HF + SiO4HF + SiO22 SiF SiF44 + 2H + 2H22OO

Example 13. Use mole ratio to convert from one reactant to the other.

0.1255 mol HF 1 mol SiO2 = 0.031 mol SiO2

4 mol HF

Compare to moles of SiO2 we have (from Step 2)

0.059 mol SiO2 (have) > 0.031 mol SiO2 (need)

4. Determine the limiting reactant•If you have more than you need, this is not the limiting reactant•This means the other reactant (HF) must be limiting

• How much excess reactant is left over?– Determine the amount of the excess

reactant used then subtract from the starting amount.

How much product can be formed?

• Start with the amount (mass or volume) of limiting reactant

• Use stoichiometry (mole ratio conversions) to find the amount (mass or volume) of product left over

Important InfoImportant Info

•Chapter 11 Homework:Chapter 11 Homework:– Red Book #23-27Red Book #23-27

• Next Class:Next Class:– S’mores Lab!S’mores Lab!– Don’t forget to sign up and Don’t forget to sign up and

bring ingredients bring ingredients

Example 2Determine the limiting reactant, if an 80.0g solution of NaOH, which is 40.0% NaOH by mass, reacts with a 75.0g solution of H2SO4, which contains 45.0% water by mass.

NaOH + H2SO4 Na2SO4 + H2O

1. Balance Equation

2. Calculate grams of NaOH and H2SO4

3. g mols of each (What you have)

4. Pick one & compare (What you need)

5. Do you have enough?

• So…the reaction will go to completion until all the NaOH (LR) is used up.

• Calculate the amounts of each product that will be produced.

ExampleExample

• If 40.0 g of HIf 40.0 g of H33POPO44 react with 60.0 g react with 60.0 g of MgCOof MgCO33 calculate: calculate:

a.a. g of Mgg of Mg33(PO(PO44))22 produced produced

b.b. g of COg of CO22 produced produced

c.c. Volume of COVolume of CO22 at STP at STP

HH33POPO44 + MgCO + MgCO33 Mg Mg33(PO(PO44))22 + CO + CO22 + + HH22OO