1. Chemical quantities and aqueous reactions
Unit 4
Chemical quantities and aqueous reactions

2. Global warming
The amount of products formed in a reaction - is related to the amount of reactants that react
Combustion of fossil fuels and climate changes - The presence of green house gases helps to control the atmospheric temperature

3. Global Warming
Green house gases allow the sunlight to enter in to the earth’s atmosphere and heat the earth’s surface
The balance between incoming and outgoing energy from sun determines the earth’s average temperature

4. Reasons for global warming
Increase of the use of fossil fuels
Increase in CO2 gas in atmosphere by burning of fuels
CO2 emitted from natural sources also contribute to global warming and its percentage is higher than fossil burning? Do you agree?

5. Reaction Stoichiometry : How much CO2 ?

6. Reaction Stoichiometry : How much CO2 ?
If we represent fossil fuel by octane, the balanced chemical equation for the combustion of octane can be represented as
C8H18 + O2  CO2 + H2O
2C8H18(l) + 25O2 (g) 16CO2(g) + 18H2O(g)
This equation gives the exact relationship between the amount of fossil fuel burned and the amount of CO2 emitted.

7. The balanced equation shows that
16 CO2 molecules are produced for every 2 molecules of octane burned
The coefficients in a chemical reaction specify the relative amounts in moles of each of substances involved in the chemical reaction

8. Reaction Stoichiometry
The numerical relationships between chemical amounts in a balanced chemical equation are called reaction stoichiometry.
Used by the chemists to plan and carry out chemical reactions to obtain products in the desired quantities.

9. Mol to Mol Relationship
After balancing equations we can now figure out the mole to mole relationships. This is also called Stoichiometric Equivalence.
For Example:
CO + 2 H2  CH3OH
1 Mol CO is equivalent to 2 mol of H2
1 mol CO is equivalent to 1 mol of CH3OH

10. The concept of Stoichiometry
1 packet of brownie mix + 1 cup vegetable oil+ 2 eggs brownies
The numerical relationship(ratio) between the ingredients
1packet of brownie mix : 20 brownies
1 cup vegetable oil : 20 brownies
2 eggs : 20 brownies

11. Combustion of octane :the numerical ratio
2 moles of C8H18 : 16 moles of CO2
We can use this ratio to determine how many moles of CO2 forms when a given number of C8H18 burns

12. Practice Problem
Consider the combustion in excess oxygen of octane.
2C8H O2  16CO2 + 18H2O
How many moles of CO2 are produced when 6.2 mol C8H18 is burned?
How many moles of oxygen are consumed in the combustion of 5.89 mol C8H18?

13. Warm-Up
Anthracene, used in the manufacture of dyes, has the mass percent composition of % C and 5.66% H. Determine is empirical formula.
Answer= C7H5

14. Gram to Gram
Lead (II) oxide reacts with ammonia as follows:
PbO + NH3  Pb + N2 + H2O
How many grams of NH3 are consumed in the reaction of g PbO?
If g Pb are produced in this reaction, how many grams of Nitrogen are also formed?

15. More Practice
The combustion of Carbon disulfide in the presence of excess oxygen yields carbon dioxide and sulfur dioxide
Write out the balance equation
The combustion of 30 g of CS2 in the presence of excess oxygen yields how many grams of SO2

16. How many moles of CO2 are produced when 0.400 mol of C6H12O6 react.
Example
The fermentation of glucose produces ethyl alcohol and CO2 in the following equation:
C6H12O6 --> 2 C2H5OH + 2 CO2
How many moles of CO2 are produced when mol of C6H12O6 react.
How many grams of C6H12O6 are needed to form 7.50g of C2H5OH?
How many grams of CO2 form when 7.50 g of C2H5OH are produced?

17. What mass of Iron (III) oxide must be used to produce 15.0g iron?
More practice
Fe2O3 + 2Al  2 Fe + Al2O3
What mass of Iron (III) oxide must be used to produce 15.0g iron?
What is the maximum mass of aluminum oxide that could be produced from 325 grams of Al

18. Comparison of CO2 produced by fossil fuel and Volcanoes
Calculate the mass of CO2 produced by the combustion of 3.5×1015 g of octane?
1.1×1016 g of CO2 (1.1×1013 kg of CO2 )per year
Volcanoes produce 2 ×1011 kg of CO2 per year

19. Limiting Reactants
Def. The reactant that is completely consumed in a chemical reaction and limits the amount of product that can be formed

20. Limiting Reactants
The limiting reactant limits the reaction and, thus, determines how much of the product forms.
The left-over reactants are called excess reactants.

21. Limiting Reactant
How can you determine which reactant in a chemical reaction is limited?

22. How to find Limiting Reactant
1st- Write out the balance chemical equation.
2nd- Find the number of moles of a reactant from each product given.
3rd-The reactant with the smallest number of grams is the limiting reactant. The one with the higher amount is the excess reactant.

23. Example # 1
What is the limiting reactant and the excess reactant when grams of Magnesium and g of Nitrogen are used to produce Magnesium Nitride.

24. Example # 2
One way to produce hydrogen sulfide gas is by the reaction of iron (II) sulfide with hydrochloric acid (HCl). If 10.2 grams of HCl is added to 13.2 g FeS, what is the limiting reactant and excess reactant.

25. Determining how much Product can be formed
1st Take the number of moles of Limiting reactant and convert to the grams of product.

26. Determining the Limiting Reactant
In the reaction below, 40.0 g of sodium hydroxide (NaOH) reacts with 60.0 g of sulfuric acid (H2SO4). How much of Sodium Sulfate can be made?

27. Determine how much excess remains
1st take the amount, usually grams, of the limiting reactant that you were giving initially and find how many grams of the excess reactant it could make in grams
2nd Subtract the beginning amount of the excess reactant by the amount that you found in step number one. This amount will be how much excess remains

28. Practice FeS + 2HCl  FeCl2 + H2S
1) One way to produce hydrogen sulfide gas is by the reaction of Iron(III) sulfide with hydrochloric acid.
FeS + 2HCl  FeCl2 + H2S
If 10.2 g HCl is added to 13.2 g FeS, how many grams of H2S can be formed? What is the mass of the excess reactant remaining?

29. Practice
Suppose that a solution containing 3.50 grams of Na3PO4 is mixed with a solution containing grams of Ba(NO3)2.
Na3PO4 + Ba(NO3)2  Ba3(PO4)2 + NaNO3
What is the limiting reactant and what is the excess reactant
How many grams of Ba3(PO4)2 can be formed?
How much of the excess reactant remains

30. Warm-Up
A 2.00 g sample of ammonia is mixed with 4.00 g of oxygen in the equation below.  What is the limiting reactant, how much water can be produced and how much excess reactant remains after the reaction has stopped?
4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(g)

31. Yield of Chemical Reaction
Theoretical Yield- the calculated quantity of product in a reaction
Actual Yield- The measured mass formed in the reaction
Note: Actual yield of chemical reactions are often less than theoretical yield

32. Is the ratio of the actual yield to the theoretical yield times 100%.
Percent Yield
Is the ratio of the actual yield to the theoretical yield times 100%.
Percent Yield = actual yield
theoretical yield
x 100 %

33. Example
If the actual yield of Mg3N2 is 47.87g and the theoretical yield is 48.85g what is the percent yield?

34. Example
Calculate the theoretical yield of ZnS, in grams, that can be made from g Zn and g S8.
8 Zn + S8  8 ZnS
If the actual yield is g ZnS, what is the percent yield?

35. Practice
Calculate the theoretical yield of C2H5Cl if 112 g of C2H5OH is reacted with 34.7 g of PCl3 based on the reaction below.  If 23.7 g of C2H5Cl is produced, what is the percent yield?
3 C2H5OH  +  PCl3  3 C2H5Cl  + H3PO3

36. Example
What mass of acetic acid is needed to prepare 252 g ethyl acetate if the expected percent yield is 85%?

37. Warm-Up 10/08/2013
In the following reaction, g of p- acetaminophenol (C6H7NO) was used to react with g of acetic anhydride (C4H6O3) to produce acetaminophen((C8H9NO2) and acetic acid (CH3COOH). The actual mass of acetaminophen produced was g. Determine the theoretical yield and the percent yield of isopentyl acetate.
C6H7NO + C4H6O3  C8H9NO2 + CH3COOH