1. Introduction to Chemical Principles Chapter 10: Chemical Calculations Involving Chemical Equations Antoine Laurent Lavoisier

2. The process by which atoms are rearranged to form different substances is called a chemical reaction. A chemical reaction is another name for a chemical change. Some reactions are hard to detect, but many provide evidence that they have occurred.

3. Evidence of Chemical Reactions A temperature change can indicate a chemical reaction. Energy released in the form of heat and light is also indicators of a chemical reaction. A color change also illustrates that a reaction has occurred. Odor, gas bubbles, and/or the appearance of a solid precipitate are also evidence that a chemical reaction has taken place.

4. Representing Chemical Reactions Chemical equations show the reaction’s reactants, which are the starting substances, and the products, which are the substances formed. Chemical equations do not express numeric quantities because the reactants are used up as the products form. Instead, the direction in which the reaction progresses is shown, which is why an arrow is used instead of an equal sign. The arrow means “yield”.

5. Symbols are used to show the physical states of the reactants and the products. See table 10-1 for the symbols used in equation writing. Word equations utilize the names of the compounds, but can be cumbersome in writing. Skeleton equations uses chemical formulas and are easier to write.

6. Writing skeleton equations is an important step, but they lack important information. The law of conservation of mass applies to each and every equation. Chemical equations must show that mass is conserved during a reaction. In order to do this, the equation must show that the number of atoms of each reactant and each product is equal on both sides.

7. The Law of Conservation of Mass Mass is neither created nor destroyed in any ordinary chemical reaction. Therefore, the total mass of reactants is always equal to the total mass of products.

8. Writing and Balancing Equations

9. In writing chemical equations, only molecular formulas are used, not empirical formulas. Reactants are written on the left side of the equation. Products are written on the right side of the equation. Reactants and products are separated by an arrow pointing to the products. Different reactants and products are separated from each other using plus signs. 2 C 3 H 7 OH + 9 O 2 6 CO 2 + 8 H 2 O

10. 2 C 3 H 7 OH + 9 O 2 6 CO 2 + 8 H 2 O Coefficients show the number of molecules or moles of each reactant and product involved in the reaction. This equation reads: “2 molecules of C 3 H 7 OH react with 9 molecules of O 2 to produce 6 molecules of CO 2 and 8 molecules of water.”

11. 2 C 3 H 7 OH + 9 O 2 6 CO 2 + 8 H 2 O In order for a chemical equation to be valid, it must: 1) Match the facts: for example, O 2, not O or O 3 is reacting with C 3 H 7 OH. 2) Obey the Law of Conservation of Mass - thus, equations must be balanced.

12. 2 Al + Fe 2 O 3 2 Fe + Al 2 O 3 Info about reaction conditions may be placed above or below the arrow. C 11 H 24 C 9 H 20 + CH 2 =CH 2  heat is added heat catalyst CH 4 + Cl 2 CH 3 Cl + HCl hv 2 H 2 O 2 H 2 + O 2 electric current light

13. (s)(s)(s)(s)(l)(l)(s)(s) The physical state of a substance may be indicated by the use of symbols in parentheses after the substance. E.g., (s) for solid (l) for liquid (g) for gas (aq) for aqueous solution 2Al + Fe 2 O 3  2Fe + Al 2 O 3

14. To balance an equation, adjust the number of atoms of each element so that they are the same on each side of the equation. Equations are balanced only by altering the equation coefficient for each item. Never, never change a substance’s formula to balance an equation.

15. Steps for Balancing Equations

16. –The formulas of the reactants and products must be correct! –The reactants are written to the left of the arrow and the products to the right of the arrow. HgO  Hg + O 2 Step 1 Write the unbalanced (skeleton) equation. Example: Write a balanced equation showing the decomposition of mercury (II) oxide to yield mercury and oxygen.

17. Step 2 Identify which elements need to be balanced. –Count and compare the number of atoms of each element on both sides of the equation. HgO → Hg + O 2 –There is one mercury atom on the reactant side and one mercury atom on the product side. –Mercury is balanced. Element Reactant Side Product Side Hg 1 1

18. Element Reactant Side Product Side O 1 2 –There are two oxygen atoms on the product side and there is one oxygen atom on the reactant side. –Oxygen needs to be balanced. HgO  Hg + O 2

19. Step 3 Balance the equation. –Balance each element one at a time, by placing whole numbers (coefficients) in front of the formulas containing the unbalanced element. –A coefficient placed before a formula multiplies every atom in the formula by that number.

20. Element Reactant Side Product Side O 1 2 Oxygen (O) is balanced. Place a 2 in front of HgO to balance O. Now there are two oxygen atoms on the reactant side and there are two oxygen atoms on the product side. HgO  Hg + O 2 2 2

21. Step 4 Check all other elements after each individual element is balanced to see whether, in balancing one element, another element became unbalanced.

22. Element Reactant Side Product Side Hg 2 1 2 HgO  Hg + O 2 There are two mercury atoms on the reactant side and there is one mercury atom on the product side. Therefore, mercury (Hg) is no longer balanced.

23. 2 HgO  Hg + O 2 Place a 2 in front of Hg to balance mercury. Mercury (Hg) is balanced. Now there are two mercury atoms on the reactant side and there are two mercury atoms on the product side. Element Reactant Side Product Side Hg 2 1 2 2

24. 2 HgO  2 Hg + O 2 Element Reactant Side Product Side Hg 2 2 O 2 2  THE EQUATION IS BALANCED 

25. General Guidelines: The coefficients in a balanced equation are the lowest whole numbers that will balance the equation. Useful to consider polyatomic ions as single entities, provided they maintain their identities in the reaction. Subscripts are never changed!!!

26. General Guidelines: It is useful, when balancing equations that have several substances, to use this approach… Balance metals. Balance nonmetals. Balance hydrogen. Balance oxygen. Subscripts are never changed!!!

27. When an aqueous solution of sulfuric acid and an aqueous solution of sodium hydroxide are mixed, the products are an aqueous solution of sodium sulfate and water. Write and Balance the Equation

28. There is one Na on the reactant side and there are two Na on the product side. Reactant Side Product Side SO 4 11 Na12 O1 1 H3 2 2 H 2 SO 4 (aq) + NaOH(aq) → Na 2 SO 4 (aq) + H 2 O(l) 2 Place a 2 in front of NaOH to balance Na. The Unbalanced (skeleton) Equation 2 4 Since the sulfate ion retains its identity in the products, we can treat it as a single entity in the balancing of the equation. The oxygen atoms not contained within the polyatomic ion will be considered separately.

29. H 2 SO 4 (aq) + NaOH(aq) → Na 2 SO 4 (aq) + H 2 O(l) There are 4 H on the reactant side and two H on the product side. Reactant Side Product Side SO 4 11 Na22 O2 1 H4 2 2 Place a 2 in front of H 2 O to balance H. 2 2 4  THE EQUATION IS BALANCED 

30. When gaseous butane (C 4 H 10 ) is completely burned in oxygen, the products are carbon dioxide and water. Write and Balance the Equation

31. C 4 H 10 (g) + O 2 (g) → CO 2 (g) + H 2 O(l) There are four C on the reactant side and there is one C on the product side. Reactant Side Product Side C 41 H102 O 2 3 4 Place a 4 in front of CO 2 to balance C. 9 4 Balance the Equation

32. C 4 H 10 (g) + O 2 (g) → CO 2 (g) + H 2 O(l) There are 10 H on the reactant side and there are two H on the product side. Reactant Side Product Side C 44 H102 O 2 9 Place a 5 in front of H 2 O to balance H. 45 10 13

33. C 4 H 10 (g) + O 2 (g) → 4 CO 2 (g) + 5 H 2 O(l) There is no whole number coefficient that can be placed in front of O 2 to balance O. Reactant Side Product Side C 4 4 H1010 O 2 13 To balance O double all of the other coefficients. 20 88 26 28 10

34. There are now 26 O on the product side. Reactant Side Product Side C 8 8 H2020 O 2 26 13 Place a 13 in front of O 2 to balance O. 26  THE EQUATION IS BALANCED  C 4 H 10 (g) + O 2 (g) → CO 2 (g) + H 2 O(l) 28 10

35.  2  6 Al + CuCl 2  Cu + AlCl 3 Al Cu Cl 1 1 1 1 2 3 2  3  6   3 33 2 Balancing Example Aluminum and copper(II) chloride react to form copper and aluminum chloride. 2

36. Types of Chemical Reactions Combustion Reactions - involve oxygen (typically from the air) Synthesis

37. A + B  AB Two reactants combine to form one product. Synthesis Reaction

38. 2Ca(s) + O 2 (g)  2CaO(s) 4Al(s) + 3O 2 (g)  2Al 2 O 3 (s) S(s) + O 2 (g)  SO 2 (g) 2Al(s) + 3Cl 2 (g)  2AlCl 3 (s) Na 2 O(s) + H 2 O(l)  2NaOH(aq) In a synthesis reaction, there is only one product.

39. AB  A + B A single substance breaks down to give two or more different substances. Decomposition Reaction

40. 2Ag 2 O(s)  4Ag(s) + O 2 (g) 2KClO 3 (s)  2KCl(s) + O 2 (g) 2NaNO 3 (s)  2NaNO 2 (s) + O 2 (g) 2H 2 O 2 (l)  2H 2 O(l) + O 2 (g) In a decomposition reaction, there is only one reactant.

41. A + BC  AC + B One element reacts with a compound to replace one of the elements of that compound. Single Displacement Reaction

42. Mg(s) + HCl(aq)  H 2 (g) + MgCl 2 (aq) Notice that the formula for the magnesium-chloride compound is MgCl 2, not MgCl. You will have to be able to write correct formulas based on your understanding of the ions involved; (Mg  Mg 2+, therefore 2 Cl - ions are required to balance the charges).

43. Metals K Ca Na Mg Al Zn Fe Ni Sn Pb H Cu Ag Hg An atom of an element in the activity series will displace an atom of any element below it from one of its compounds. Sodium (Na) will displace any element below it from one of its compounds. increasing activity The Activity Series

44. Mg(s) + PbS(s)  MgS(s) + Pb(s) Magnesium is above lead in the activity series. Metals Mg Al Zn Fe Ni Sn Pb

45. Ag(s) + CuCl 2 (s)  no reaction ? Metals Pb H Cu Ag Hg Silver is below copper in the activity series.

46. Cl 2 (g) + CaBr 2 (s)  CaCl 2 (aq) + Br 2 (aq) There is also a halogen activity series – it has the same order as the halogens are found in the Periodic Table Halogens F 2 Cl 2 Br 2 I 2 Chlorine is above bromine in the activity series. ?

47. AB + CD  AD + BC Two compounds exchange partners with each other to produce two different compounds. The reaction can be thought of as an exchange of positive and negative groups. A combines with D Double Displacement Reaction And B combines with C

48. formation of a precipitate formation of a gas (release of bubbles) release or absorption of heat formation of a molecular compound (e.g., H 2 O) One or more of the following will accompany a Double Displacement Reaction A precipitate is a reaction product that isn’t soluble in the reaction medium; it falls out of solution as it forms.

49. Acid Base Neutralization HCl(aq) + NaOH(aq)  NaCl(aq) + H 2 O(l) H 2 SO 4 (aq) + 2NaOH(aq)  Na 2 SO(aq) + 2H 2 O(l) acid + base → salt + water

50. Formation of an Insoluble Precipitate AgNO 3 (aq) + NaCl(aq)  AgCl(s) + NaNO 3 (aq) Pb(NO 3 ) 2 (aq) + 2KI(aq)  PbI 2 (s) + 2KNO 3 (aq)

51. Metal Oxide + Acid CuO(s) + 2HNO 3 (aq)  Cu(NO 3 ) 2 (aq) + H 2 O(l) CaO(s) + 2HCl(aq)  CaCl 2 (s) + H 2 O(l) metal oxide + acid → salt + water

52. Formation of a Gas H 2 SO 4 (aq) + 2NaCN(aq)  Na 2 SO 4 (aq) + 2HCN(g) NH 4 Cl(aq) + NaOH(aq)  NaCl(aq) + NH 4 OH(aq) NH 4 OH(aq)  NH 3 (g) + H 2 O(l) indirect gas formation

53. Na(s) + O 2 (g)  4 2 Combustion Reaction Products: –contain oxygen –hydrocarbons (C x H y ) form CO 2 + H 2 O Na 2 O(s) CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O(g) A reaction in which a substance reacts with oxygen and which proceeds with the evolution of heat. A + O 2  B

54. Proportional Relationships I have 5 eggs. Assuming I have plenty of the other ingredients, how many cookies can I make? 3/4 c. brown sugar 1 tsp vanilla extract 2 eggs 2 c. chocolate chips Makes 5 dozen cookies. 2 1/4 c. flour 1 tsp. baking soda 1 tsp. salt 1 c. butter 3/4 c. sugar 5 eggs 5 dozen cookies 2 eggs = 12.5 dozen cookies Ratio of eggs to cookies

55. Chemical Equations and the Mole P 4 O 10 + 6 H 2 O  4 H 3 PO 4 1 molecule of P 4 O 10 reacts with 6 molecules of H 2 O to give 4 molecules of H 3 PO 4 1 mole of P 4 O 10 reacts with 6 moles of H 2 O to give 4 moles of H 3 PO 4

56. Stoichiometry and Chemical Equations The coefficients in a chemical equation not only describe the number of atoms/molecules being reacted/produced, they also describe the number of moles of atoms/molecules being reacted/produced. N 2 (g) + 3H 2 (g) 2NH 3 (g) Suppose 5.46 mol N 2 completely react in this fashion: 5.46 mol N 2 3 mol H 2 1 mol N 2 = 16.4 mol H 2 are needed to react 5.46 mol N 2 2 mol NH 3 1 mol N 2 = 10.9 mol NH 3 must be produced

57. Stoichiometry Practice Bismuth(III) chloride will react with hydrogen sulfide to form bismuth(III) sulfide and hydrochloric acid. Write the balanced equation for this reaction, then calculate how many moles of acid would be formed if 15.0 mol of hydrogen sulfide react. 2 BiCl 3 + 3 H 2 S2 Bi 2 S 3 + 6 HCl 15.0 mol H 2 S 6 mol HCl 3 mol H 2 S = 30.0 mol HCl

58. Stoichiometry Practice Sodium nitride can be formed by reacting sodium metal with nitrogen gas. Write the balanced equation for this reaction, then calculate how many moles of sodium nitride can be produced from 25.0 g of sodium. 6 Na + N 2 2 Na 3 N 25.0 g Na1 mol Na2 mol Na 3 N 22.99 g Na6 mol Na = 0.362 mol Na 3 N

59. How many bicycles can be assembled from the parts shown? From eight wheels four bikes can be constructed. From four frames four bikes can be constructed. From three pedal assemblies three bikes can be constructed. The limiting part is the number of pedal assemblies.

60. Likewise, in a chemical reaction, it is possible for one reactant to be used up before the other(s). A limiting reactant is a substance that is completely used up in a chemical reaction and causes the reaction to stop. Therefore, it is the limiting reactant that also determines how much product can actually be formed.

61. To determine which reactant is limiting in a given reaction: Determine the number of moles present of each reactant. Calculate the moles of product if a particular reactant were to be completely used up. Do this for all reactants. The reactant that produces the fewest number of moles of product is the limiting reactant.

62. Limiting Reactant Example Nitrogen gas and hydrogen gas can react to form gaseous ammonia. How many moles of ammonia can be prepared from the reaction of 64.7 g of nitrogen and 12.1 g of hydrogen? N 2 + 3 H 2 2 NH 3 64.7 g N 2 1 mol N 2 2 mol NH 3 28.02 g N 2 1 mol N 2 = 4.62 mol NH 3 12.1 g H 2 1 mol H 2 2 mol NH 3 2.02 g H 2 3 mol H 2 = 3.99 mol NH 3 smallest H 2 is limiting

63. Limiting Reactant Practice Sodium metal will react with gaseous ammonia to produce solid sodium amide, NaNH 2. The unbalanced equation for this reaction is… Na + NH 3 NaNH 2 + H 2 If 60.0 g of sodium are mixed with 48.0 g of ammonia, how many grams of sodium amide can be produced? Which substance is the limiting reactant?

64. Limiting Reactant Practice Na + NH 3 NaNH 2 + H 2 60.0 g Na1 mol Na2 mol NaNH 2 22.99 g Na2 mol Na = 2.61 mol NaNH 2 48.0 g NH 3 1 mol NH 3 2 mol NaNH 2 17.04 g NH 3 3 mol NH 3 = 2.82 mol NaNH 2 smallest Na is limiting 222

65. Theoretical, Actual & Percent Yields Theoretical yield: the amount of product you would get if the reaction occurs with complete efficiency. Actual yield: what you “actually” get when you perform the reaction. Percent yield: the actual yield divided by the theoretical yield multiplied by 100. % Yield = (Actual/Theoretical) x 100

66. Percent Yield calculated on paper measured in lab

67. When 45.8 g of K 2 CO 3 react with excess HCl, 46.3 g of KCl are formed. Calculate the theoretical and % yields of KCl. K 2 CO 3 + 2HCl  2KCl + H 2 O + CO 2 45.8 g? g actual: 46.3 g

68. 45.8 g K 2 CO 3 1 mol K 2 CO 3 138.21 g K 2 CO 3 = 49.4 g KCl 2 mol KCl 1 mol K 2 CO 3 74.55 g KCl 1 mol KCl K 2 CO 3 + 2HCl  2KCl + H 2 O + CO 2 45.8 g? g actual: 46.3 g Theoretical Yield:

69. Theoretical Yield = 49.4 g KCl % Yield = 46.3 g 49.4 g  100 = 93.7% K 2 CO 3 + 2HCl  2KCl + H 2 O + CO 2 45.8 g 49.4 g actual: 46.3 g