PRODUCTS WHY CAN’T I USE ALL MY BUILDING BLOCKS? (REACTANTS) WE DON’T HAVE ENOUGH OF EACH OF THE...

PRODUCTS

WHY CAN’T I USE ALL MY BUILDING

BLOCKS? (REACTANTS)

•WE DON’T HAVE ENOUGH OF EACH OF THE BLOCKS TO BUILD FULL PRODUCTS

•SOME REACTANTS ARE LEFT OVER BUT ONE REACTANT GETS COMPLETELY USED UP!

Limiting Reagents

STOICHIOMETRY

•the the reactantreactant which is totally which is totally consumed when the chemical consumed when the chemical reaction is completereaction is complete

Limiting Reagent

“the amount of product formed is limited by this reactant”

Limiting reagentLimiting reagent

Q - How many moles of NO are produced if 4 mol NH3 are burned in 5 mol O2?

Given: 4NH3 + 5O2 6H2O + 4NO

4 mol NO, works out PERFECTLY – both reactants are completely used up

Here, NH3 limits the production of NO; If there was more NH3, more NO would be produced

NH3 is called the “limiting reagent” and O2 is in “excess

4 mol NO, with leftover O2

Given: 4NH3 + 5O2 6H2O + 4NO

Q - How many moles of NO are produced if 4 mol NH3 are burned in 20 mol O2?

On your worksheet: How many moles of NO are produced if

4 mol NH3 are burned in 2.5 mol O2?

2 mol NO, with leftover NH3

• Here, O2 limits the production of NO; if there was more O2, more NO would be produced

• Thus, O2 is called the “limiting reagent” and NH3 is in excess!

How can I tell which reactant is the limiting

reagent?Use a comparison chart between what

we have and what the balanced equation says we need…

NH3 O2

What we have Mole given Mole given

From the question

Mole ratio calculated

What we need Ratio in balanced equation

Ok… let’s try it!!4NH3 + 5O2 6H2O + 4NO

3.2 mol NH3 reacts with 1.6 mol O2

-which reactant will limit the production of the reactants?

NH3 O2

What we have

What we need

Comparison chartComparison chart

3.2 1.6

3.2/1.6 = 2 mol

1.6/1.6 = 1 mol

4/5 = 0.8 5/5 = 1

There is more NH3 than needed to react all the O2.

So O2 is the limiting reagent which makes NH3 the

excess reagent!• Now you can use the limiting

reagent moles to calculate how much product you can make!

Limiting reagents in stoichiometryLimiting reagents in stoichiometry

How many moles of NO are produced if 0.25 moles NH3 are burned in 0.56 mol O2? (make a chart)

4NH3 + 5O2 6H2O + 4NO

4 mol NO 4 mol NH3

x

NO

mol= 0.25 mol NH3 =0.25 mol NO

0.25 mol NH3 is the limiting reagent

Complete questions 3 and 4 on the worksheet!

Al(s) + MnO2(aq) Al2O3(aq) + Mn(s)

3. What is the limiting reactant when 0.1372 mol of aluminum reacts with 0.1264 mol of MnO2? How many moles of the aluminum product should be yielded from this reaction?

Al MnO2

What we have

What we need

0.1372 0.1264

0.1372/0.1264

1.085

0.1264/0.1264

1

3 4 2 3

4/3 = 1.33 3/3 = 1

So, based on the balanced equation, to use all the MnO2 I would need

more Aluminum than I have?

Aluminum is the LIMITING REACTANT!

Mol Al2O3 made = 0.1372 mol Al x 2 mol Al2O3

4 mol Al = 0.0686 mol

4. If 0.434 mol of both reactants are combined, which will be the limiting reagent?

What is the theoretical (predicted) maximum mass of manganese that can be yielded from this reaction?

LIMITING REACTANT IS aluminum

Theoretical maximum of Mn is 0.3255 mol which (using mm of Mn) converts to 17.88 g

•Sometimes the question is more complicated. For example, if grams of the two reactants are given instead of moles we must first determine moles, then decide which is limiting …

Solving Limiting reagents mass to moleSolving Limiting reagents mass to mole

Q - How many g NO are produced if 20 g NH3 is burned in 30 g O2?

A - First we need to calculate the number of moles of each reactant

4NH3 + 5O2 6H2O + 4NO

1 mol NH3 17.0 g NH3

x # mol NH3= 20 g NH3 1.176 mol NH3

=

1 mol O2 32.0 g O2

x # mol O2= 30 g O20.9375 mol O2

=

NH3 O2

What we have

What we need

1.176 0.937

1.176/0.937 = 1.25 mol

0.937/0.937 = 1 mol

*Choose the smallest value to divide each by

4 5

Comparison chartComparison chart

A – Once the number of moles of each is calculated-find the LR…

A - There is more NH3 (what we have) than needed (what we need). Thus NH3 is in excess, and O2 is the limiting reagent.

StoichiometryStoichiometry1) Expressed all chemical quantities as moles

2) Determined the limiting reagent via a chart 3) Use the limiting reagent to determine how

much product can be made

Limiting Reagents: “shortcut”Limiting Reagents: “shortcut”• Limiting reagent problems can be solved

another way (without using a chart)…• Do two separate calculations using both given

quantities. The smaller answer is correct.Q - How many g NO are produced if 20 g NH3 is

burned in 30 g O2? 4NH3 + 5O2 6H2O+ 4NO

4 mol NO5 mol O2

x 30 g O2

22.5 g NO=

30.0 g NO1 mol NO

x 1 mol O2 32.0 g O2

x

4 mol NO4 mol NH3

x

# g NO=20 g NH3

35.3 g NO=

30.0 g NO1 mol NO

x 1 mol NH3 17.0 g NH3

x

Practice questionsPractice questions1. 2Al + 6HCl 2AlCl3 + 3H2

If 25 g of aluminum was added to 90 g of HCl, what mass of H2 will be produced (try this two ways – with a chart & using the shortcut)?

2. N2 + 3H2 2NH3: If you have 20 g of N2 and 5.0 g of H2, which is the limiting reagent?

3. What mass of aluminum oxide is formed when 10.0 g of Al is burned in 20.0 g of O2?

4. When C3H8 burns in oxygen, CO2 and H2O are produced. If 15.0 g of C3H8 reacts with 60.0 g of O2, how much CO2 is produced?

5. How can you tell if a question is a limiting reagent question vs. typical stoichiometry?

6. If 25 g magnesium chloride was added to 68 g silver nitrate, what mass of AgCl will be produced?

MgCl2 + 2AgNO3 Mg(NO3)2 + 2AgCl

11 1 mol Al 27.0 g Al

x # mol Al =25 g Al = 0.926 mol

# mol HCl = 90 g HCl 1 mol HCl 36.5 g HCl

x = 2.466 mol

Al HCl

What we

have

What we

need

0.926 2.466 0.926/0.926

= 1 mol 2.466/0.926

= 2.7 mol

2 6 2/2 = 1 mol 6/2 = 3 mol

HCl is limiting.

3 mol H2

6 mol HClx

# g H2 =90 g HCl 2.0 g H2

1 mol H2

x 1 mol HCl 36.5 g HCl

x = 2.47 g H2

Question 1: shortcutQuestion 1: shortcut

2Al + 6HCl 2AlCl3 + 3H2

If 25.0 g aluminum was added to 90.0 g HCl, what mass of H2 will be produced?

3 mol H2

2 mol Al x # g H2= 25 g Al = 2.78 g H2

2.0 g H2

1 mol H2

x 1 mol Al27.0 g Al

x

3 mol H2

6 mol HClx # g H2 = 90 g HCl = 2.47 g H2

2.0 g H2

1 mol H2

x 1 mol HCl36.5 g HCl

x

N2 H2

What we have

What we need

Question 2Question 2

0.714 mol 2.5 mol

0.714/0.714 = 1 mol

2.5/0.714 = 3.5 mol

We have more H2 than what we need, thus H2 is in excess and N2 is the limiting factor.

1 mol 3 mol

1 mol N2 28 g N2

x # mol N2= 20 g N2 0.714 mol N2=

1 mol H2 2 g H2

x # mol H2= 5.0 g H2 2.5 mol H2=

Question 2: shortcutQuestion 2: shortcut

N2 + 3H2 2NH3

If you have 20 g of N2 and 5.0 g of H2, which is the limiting reagent?

2 mol NH3

1 mol N2 x

# g NH3=

20 g N2 = 24.3 g H2 17.0 g NH3

1 mol NH3

x 1 mol N2

28.0 g N2

x

2 mol NH3

3 mol H2

x

# g NH3 =5.0 g H2 = 28.3 g H2

17.0 g NH3

1 mol NH3

x 1 mol H2

2.0 g H2

x

N2 is the limiting reagent

Al O2

33 4Al + 3O2 2 Al2O3

1 mol Al 27 g Al

x # mol Al = 10 g Al 0.37 mol Al=

1 mol O2 32 g O2

x # mol O2 = 20 g O20.625 mol O2=

0.37 mol 0.625 mol 0.37/.37 =

1 mol 0.625/0.37

= 1.68 mol

4 mol 3 mol 4/4 = 1 mol 3/4 = 0.75 mol

What we have

What we need

There is more than enough O2; Al is limiting

2 mol Al2O3

4 mol Al x # g Al2O3 = 0.37 mol Al

18.9 g Al2O3=

102 g Al2O3

1 mol Al2O3

x

Question 3: shortcutQuestion 3: shortcut

4Al + 3O2 2 Al2O3

What mass of aluminum oxide is formed when 10.0 g of Al is burned in 20.0 g of O2?

2 mol Al2O3

4 mol Al x

# g Al2O3=

10.0 g Al = 18.9 g Al2O3 102.0 g Al2O3

1 mol H2

x 1 mol Al27.0 g Al

x

2 mol Al2O3

3 mol O2

x

# g Al2O3=

20.0 g O2 = 42.5 g Al2O3 102.0 g Al2O3

1 mol H2

x 1 mol O2

32.0 g O2

x

C3H8 O2

44 C3H8 + 5O2 3CO2 + 4H2O1 mol C3H8 44 g C3H8

x # mol C3H8 = 15 g C3H80.34 mol

C3H8

=

1 mol O2 32 g O2

x # mol O2 = 60 g O21.875 mol O2=

0.34 mol 1.875 mol 0.34/.34 = 1

mol 1.875/0.34

= 5.5 mol

1 mol 5 mol

What we have

Need

3 mol CO2

1 mol C3H8 x

# g CO2 =0.34 mol C3H8

45.0 g CO2=

44 g CO2

1 mol CO2

x

We have more than enough O2, C3H8 is limiting

Question 4: shortcutQuestion 4: shortcutC3H8 + 5O2 3CO2 + 4H2O

When C3H8 burns in oxygen, CO2 and H2O are produced. If 15.0 g of C3H8 reacts with 60.0 g

of O2, how much CO2 is produced?3 mol CO2

1 mol C3H8 x

# g CO2=15.0 g C3H8 = 45.0 g CO2

44.0 g CO2

1 mol CO2

x 1 mol C3H8

44.0 g C3H8

x

3 mol CO2

5 mol O2

x

# g CO2=60.0 g O2 = 49.5 g CO2

44.0 g CO2

1 mol CO2

x 1 mol O2

32.0 g O2

x

5. Limiting reagent questions give values for two or more reagents (not just one)

Question 6: shortcutQuestion 6: shortcutMgCl2 + 2AgNO3 Mg(NO3)2 + 2AgCl

If 25.00 g magnesium chloride was added to 68.00 g silver nitrate, what mass of AgCl will be produced?

2 mol AgCl1 mol MgCl2

x

# g AgCl=25 g MgCl2

75.25 g AgCl=

143.3 g AgCl1 mol AgCl

x 1 mol MgCl295.21 g MgCl2

x

2 mol AgCl2 mol AgNO3

x

# g AgCl=68 g AgNO3

57.36 g AgCl=

143.3 g AgCl1 mol AgCl

x 1 mol AgNO3

169.88 g AgNO3

x