Chapter 11: Stoichiometry: Calculations Based on Chemical Equations

1

• Chapter 11 Stoichiometry Calculations Based on
  Chemical Equations

2
Stoichiometry

• Stoichiometry is the calculation of quantities of
  substances involved in chemical reactions.
• Balanced chemical equations, through their
  coefficients indicate the ratio of moles of
  reactants and products

3
N2 3H2 2NH3

• Mole ratios

1 mol N2 3 mol H2
3 mol H2 1 mol N2
or
2 mol NH3 1 mol N2
1 mol N2 2 mol NH3
or
2 mol NH3 3 mol H2
3 mol H2 2 mol NH3
or
4
Mole-Mole Calculations

• Given the moles of one reactant or product,
  calculate the moles required or produced in the
  other reactant(s) or product(s).
• Steps
• 1. Write the balanced chemical equation.
• 2. Write the moles of the given reactant or
  product.
• 3. Multiply by the mole ratio
• (moles desired)/(moles starting chemical)

5
What does this equation mean?

• N2 3H2 2NH3

3 molecules of hydrogen (each containing 2 atoms)
to form
2 molecules of ammonia. ( Each molecule contains
1 atom of nitrogen and 3 atoms of hydrogen.)
1 molecule of nitrogen (each containing 2 atoms)
reacts with
3 moles of hydrogen (H2) to form
1 mole of nitrogen (N2) reacts with
2 moles of ammonia (NH3)
6
Convert moles to moles

• moles wanted
• moles known( coef wanted/coef known)
• or
• Moles of desired chemical

Moles of desired chemical Moles of starting
chemical
Given moles of starting chemical
x
7
For the reaction equationaA bB cC
dD

• Mole B mol A x (b/a)
• Mole C mol A x (c/a)
• Mole D mol A x (d/a)

8
N2 3H2 2NH3

• Calculate the moles of NH3 that can be produced
  from 10.8 moles of H2.
• Mole C mol A x (c/a)
• Mol NH3 mol H2 x (coef NH3)/(coef H2)
• mol NH3 10.8 mol H2 ( 2 NH3)/(3 H2)

Mol NH3 7.20 mol NH3
9
Steps in doing stoichiometry

• Converting Grams of A to Grams of B
• 1. Convert grams to moles
• 2. Convert moles to moles
• 3. Convert moles to grams

10
Convert grams known to moles

• Moles g/MM

11
Convert moles to moles

• moles wanted
• moles known( coef wanted/coef known)

12
Convert moles to grams

• g mole x MM

13
Route to Calculating Mass Relationships
Insert figure 11.1
14
Stoichiometric Process in One Equation

• For the equation aA bB

Grams A Molar Mass A
b a
Grams B
X
X (Molar Mass B)
15
Calculating Mass of Unknown Quantities from Known
Quantities
16
Example Problem

• N2 3 H2 2NH3
• Start with 50 g N2. How many grams of NH3 will
  be produced?
• MM N2 2 x 14 28 g
• MM NH3 14 3x1 17 g
• mol N2 g/MM 50/28 1.78 mol
• mol NH3 1.78 mol N2 x (2 NH3/1 N2) 3.56 mol
• g NH3 mol x MM 3.56x 17 60.53 g

17
Moles to Grams

• In doing a Stoichiometry calculation, you can
  start and end at any point in the process,
  depending upon what is given and what is desired
• Grams A moles A moles B grams B
• To Convert from moles A to grams B, follow the
  sequence
• moles A moles B grams B

18
N2 3 H2 2NH3

• Calculate the grams of N2 that will react with
  12.0 moles of H2.
• Moles N2 (12.0 moles H2)x(1 N2/3 H2)
• 4.00 moles N2
• Molar Mass N2 2(14.0) 28.0 g/mol
• grams N2 4.00 moles x 28.0 g/mol
• 112 g N2

19
Calculations Involving Molar Solutions

• Calculations of the volumes of solutions can be
  made by modifying the stoichiometry process
  slightly
• 1. Convert grams or volume of A to moles A
• 2. Convert moles A to moles B
• 3. Convert moles B to volume B or grams B
• Remember Molarity Moles/Liters
• Moles Liters x Molarity
• Liters Moles/Molarity

20
Stoichiometry Calculations Involving Solutions
Insert figure 11.2
21
Stoichiometry with Volumes

• Calculate the volume of 0.987 M NaOH required to
  neutralized 50 mL of 0.670 M HCl
• NaOH(aq) HCl(aq) NaCl(aq) H2O(l)
• mol HCl MxL 0.670(0.050) 0.0335 mol
• mol NaOH mol HCl 0.0335 mol (from the
  equation)
• L NaOH mol/M 0.0335/.987 0.0339 L
• 33.9 mL NaOH.

22
Limiting Reactants

• In a chemical reaction, there is an exact ratio
  of reactants, determined by the balanced chemical
  equation.
• If there is too much of one of the reactants, it
  is said to be in excess.
• The other reactant is said to be the limiting
  reactant (or reagent).

23
A Limiting Reagent Analogy
24
Molecular Model of Limiting Reagents
25
Steps in Identifying the Limiting Reagent

• 1. Determine the moles of each reactant
  (molg/MM)
• 2. Divide the moles of each reactant by its
  coefficient.
• 3. The lesser of the amounts in step 2 is the
  limiting reagent. The other reactants are said
  to be in excess.

26
Example LR problem

• N2 3H2 2 NH3
• If you start with 50.0g of Nitrogen and 45.0g of
  Hydrogen, Determine which reactant is in Excess
  and which is Limiting.
• Moles N2 50/28 1.78
• Moles H2 45/2 22.5
• Divide by coeff. 1.78/1 1.78 22.5/3 7.5
• nitrogen is less, LR, hydrogen is in excess.

27

Calculate theoretical, actual and percentage
yield, given appropriate data.

• The theoretical yield would be the amount of
  product you would predict to be produced from a
  given quantity of reactant(s) from stoichiometric
  calculations.
• The actual yield is the amount of product
  actually obtained by weighing.
• yield (actual/theoretical) x 100

28
Example Problem

• When 5.00g of N2 are reacted with 7.00g of H2,
  5.08g of NH3 are produced. Calculate the
  theoretical and the percentage yield.
• The actual yield is 5.08g.
• The balanced equation is
• N2(g) 3H2(g) 2NH3(g)
• 1. Determine the Limiting Reagent
• a. Determine the moles of each reactant.
• Mol g/M M(N2) 2x14 28

29
When 5.00g of N2 are reacted with 7.00g of H2,
5.08g of NH3 are produced. Calculate the
theoretical and the percentage yield.

• N2(g) 3H2(g) 2NH3(g)
• M(H2) 2x1 2
• mol N2 5.00/28 0.179 mol
• mol H2 7.00/2 3.5 mol
• b. Divide each of the mol by its coefficient
• N2 0.179/1 0.179
• H2 3.5/3 1.17
• c. Choose the smaller of the answers in b. as
  the Limiting Reagent N2 Base further
  calculations on LR N2

30
When 5.00g of N2 are reacted with 7.00g of H2,
5.08g of NH3 are produced. Calculate the
theoretical and the percentage yield.

• N2(g) 3H2(g) 2NH3(g)
• 2. Calculate theoretical yield based on the LR,
  N2
• a. Mol N2 0.179 mol (see previous calc.)
• b. Mol NH3 mol N2x (coef. NH3/coef. N2)
• 0.179x(2/1) 0.358 mol
• c. mass NH3 molxM
• M 14 3(1) 17
• mass NH3 0.358 x 17 6.09 g Th. Yield

31
Percentage Yield

• Percentage Yield (Actual/Theoretical)x100
• (5.08/6.09)x100
• 83.4

32
Energy Changes in Chemical Reactions

• Many reactions give off heat energy.
• This energy is measured in Joules (J), kilojoules
  (kJ), calories (cal), or kilocalories (kcal)
• 1 cal 4.184 J
• 1 kcal 4.184 kJ
• The heat energy given up may be treated as a
  product In the combustion of methane
• CH4(g) 2 O2(g) CO2(g) H2O(g) heat

33
Factors in Heat Energy

• Chemicals present
• Amounts used
• Conditions of reaction
• At 1 atm pressure, 25oC, for 1 mol CH4
• CH4(g) 2 O2(g) CO2(g) H2O(g) 802 kJ
• CH4(g) 2 O2(g) CO2(g) H2O(l) 890 kJ
• More heat is given of if H2O is produced in the
  liquid state compared to the gas state.

34
Different Types of Energy May be Given Off or
Absorbed

• If heat energy is given off, the reaction is
  exothermic.
• If other types of energy are given off, the
  reaction is exergonic.
• If heat energy is absorbed, the reaction is
  endothermic.
• If other types of energy is absorbed, the
  reaction is endergonic.

35
Enthalpy

• The amount of energy given off in a chemical
  reaction is equal to the difference in chemical
  energy between the reactants and products. When
  pressure is constant, this energy change is
  called the enthalpy of the reaction, DH.
• The reaction of hydrogen with oxygen to form
  water is exothermic
• H2(g) ½O2(g) H2O(l) 283 kJ
• This can also be written
• H2(g) ½O2(g) H2O(l) DH -283 kJ
• Exothermic reactions have negative DH values.

36
Endothermic Reactions

• The reverse of this reaction corresponds to the
  electrolysis of water.
• H2O(l) 283 kJ H2(g) ½O2(g)
• H2O(l) H2(g) ½O2(g) DH 283 kJ
• Endothermic Reactions have a positive enthalpy
  change.
• The reverse reaction has an enthalpy change of
  the same magnitude but opposite sign.

37
Quantitative Considerations

• H2(g) ½O2(g) H2O(l) DH -283 kJ
• Calculate the heat given off when 5.0 g of water
  is produced
• The heat is proportional to the number of moles.
  The above reaction is for 1 mol.
• 5.0 g/ 18g/mol 0.277 mol
• heat given off
• (-283 kJ/mol) x 0.277mol 78.6 kJ

38
Tendency for Reactions to Occur

• First Law of Thermodynamics
• Energy is conserved. In giving off energy,
  compounds go to a lower energy state. Systems
  tend to move to lower energy states.
• Endothermic reactions can occur because systems
  also tend to become more random.
• This randomness is called entropy. Change in
  entropy is DS.
• According to the Second Law of Thermodynamics,
  the entropy of the universe is always increasing.
  If in a chemical reaction the entropy of the
  universe increases, the reaction is spontaneous.