Unit 6: Concept 3

Limiting Reactants

• You can only make as much product as the

reactants you have.

• Once one reactant is used up, no more product

can be formed (even if you still have the other

reactant!)

– Limiting reactant = the reactant that limits the

amount of product that can be made

• It gets used up first, and thus can make the least

amount of product

– Excess reactant = the reactant that isn’t

completely used up in the reaction

• Thus you will have excess or “leftovers” of this one

Limiting Reactants

How to identify limiting vs. excess

Example: Consider the following reaction.

H4(l) + 2H2

4.50 mol of H2

1 mol N2

2 mol H2

O2(l)  N2

a. If you have 6.00 mol of N2

6.00 mol N2

4.50 mol H2

(g) + 4H2

H2O2, which reactant is limiting?

6.00 mol of N2

H4 = 1 mol N2

O2 = 1 mol N2

1 mol N2

O(g)

H4 and 4.50 mol of

= 6.00 mol N2

1 mol N2

2 mol H2

= 2.25 mol N2

∴ H2O2 is the limiting

reactant

Limiting Reactants

How to identify limiting vs. excess

Example: Consider the following reaction.

H4(l) + 2H2

6.00 mol of N2

4.50 mol of H2

2.25 mol N2

1 mol N2

made

H4 = 1 mol N2

6.00 mol – 2.25 mol = 3.75 mol N2

unused

O2(l)  N2

2.25 mol N2

(g) + 4H2

1 mol N2

O(g)

b. How much of the excess reactant remains

unused?

2.25 mol

N2H4 needed

Limiting Reactants

How to identify limiting vs. excess

Example: Consider the following reaction.

H4(l) + 2H2

O2(l)  N2

(g) + 4H2

2.25 mol N2 (because that’s how much the limiting reactant H2

4.50 mol of H2

2 mol H2

mol H2

O = ?

= 9.00 mol H2

O2 = 4 mol H2

4.50 mol H2

4 mol H2

2 mol H2

O(g)

c. How much of each product, in moles, is

formed?

O2 can make)

18.0 mol H2

Practice Time!

1. Consider the following reaction to answer the

questions below:

SiO2

(s) + 4HF(g)  SiF4

(g) + 2H2

and 0.500 mol of HF.

O(l)

a. Identify the limiting reactant if you begin with

0.750 mol of SiO2

b. How much of the excess reactant, in moles, will

be leftover?

c. How much of each product, in moles, is

formed?

d. How much of each product, in grams, is

formed?

Percent Yield

• Actual yield: the measured amount of product

actually made from a reaction

• Theoretical yield: the maximum amount of product

that could be made from a given amount of reactant

• Percent yield: the ratio of what could have been

produced to what actually was produced

Percent yield =

Actual yield

Theoretical yield

x 100

Percent Yield

Example: Consider the reaction of photosynthesis

below. 45.7 g of H2

CO2 and the actual yield is 32.8 g of C6

is the percent yield of glucose (C6

6CO2

+ 6H2

H12O6

= ?

O  C6

H12O6

45.7 g H2O = limiting reactant Molar mass of H2

32.8 g C6H12O6 = actual yield

theoretical yield of C6

percent yield of C6

6 mol H2

H12O6

O = 1 mol C6

= ?

H12O6

1(16.00)

Molar mass of H2

45.7 g H2

H12O6)?

+ 6O2

O = 2(1.008) +

O = 2.016 + 16.00

O = 18.016 g/mol

1 mol H2

18.016 g H2

2.54

mol H2

O are reacted with an excess of

H12O6. What

Percent Yield

Example: Consider the reaction of photosynthesis below. 45.7 g

of H2

O are reacted with an excess of CO2

H12O6

6CO2

+ 6H2

O  C6

H12O6

and the actual yield is

32.8 g of C6H12O6. What is the percent yield of glucose

(C6

+ 6O2

45.7 g H2O = limiting reactant Molar mass of C6

32.8 g C6H12O6 = actual yield

theoretical yield of C6

percent yield of C6

6 mol H2

H12O6

2.54 mol H2

O = 1 mol C6

H12O6

= ?

12.096 + 96.00

Molar mass of C6

H12O6 = 6(12.01) +

12(1.008) + 6(16.00)

Molar mass of C6

H12O6

= 72.06 +

H12O6 = 180.16 g/mol

Percent Yield

Example: Consider the reaction of photosynthesis below. 45.7 g

of H2

O are reacted with an excess of CO2

H12O6

6CO2

2.50 mol H2

O 1 mol C6

+ 6H2

H12O6

6 mol H2

Actual yield

Percent yield =

x 100

Theoretical yield

32.8 g C6

75.07 g C6

H12O6

x 100 = 43.7%

O  C6

H12O6

180.16 g C6

1 mol C6

H12O6

and the actual yield is

32.8 g of C6H12O6. What is the percent yield of glucose

(C6

+ 6O2

H12O6

450.4 g C6

H12O6

= 75.07 g C6

H12O6

(theoretical yield)

Practice Time!

Percent Yield

1. Find the percent yield of NH3

are reacted with excess H2

NH3

are actually made.

+ 3H2

if 25.20 g of N2

and 29.98 g of

 2NH3