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?