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Introduction to Chemical Principles

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

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Introduction to Chemical Principles

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

  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 andBalancing Equations

  9. In writing chemical equations, only molecular formulas are used, not empirical formulas. 2 C3H7OH + 9 O2 6 CO2 + 8 H2O Products are written on the right side of the equation. Reactants are written on the left 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.

  10. 2 C3H7OH + 9 O2 6 CO2 + 8 H2O Coefficients show the number of molecules or moles of each reactant and product involved in the reaction. This equation reads: “2 molecules of C3H7OH react with 9 molecules of O2 to produce 6 molecules of CO2 and 8 molecules of water.”

  11. 2 C3H7OH + 9 O2 6 CO2 + 8 H2O In order for a chemical equation to be valid, it must: 1) Match the facts: for example, O2, not O or O3 is reacting with C3H7OH. 2) Obey the Law of Conservation of Mass - thus, equations must be balanced.

  12. Info about reaction conditions may be placed above or below the arrow. heat is added  2 Al + Fe2O3 2 Fe + Al2O3 heat C11H24 C9H20 + CH2=CH2 catalyst electric 2 H2O 2 H2 + O2 current light hv CH4 + Cl2 CH3Cl + HCl

  13. 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 + Fe2O3  2Fe + Al2O3 (s) (s) (l) (s)

  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. Example: Write a balanced equation showing the decomposition of mercury (II) oxide to yield mercury and oxygen. HgO  Hg + O2 Step 1Write the unbalanced (skeleton) equation. • 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.

  17. Step 2Identify which elements need to be balanced. HgO → Hg + O2 • Count and compare the number of atoms of each element on both sides of the equation. Element Reactant Side Product Side Hg 1 1 • There is one mercury atom on the reactant side and one mercury atom on the product side. • Mercury is balanced.

  18. HgO  Hg + O2 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.

  19. Step 3Balance 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. 2 2 HgO  Hg + O2 • Place a 2 in front of HgO to balance O. Element Reactant Side Product Side O 1 2 • Now there are two oxygen atoms on the reactant side and there are two oxygen atoms on the product side. • Oxygen (O) is balanced.

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

  22. 2 HgO  Hg + O2 • 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. Element Reactant Side Product Side Hg 2 1

  23. 2 2 2 HgO  Hg + O2 Element Reactant Side Product Side Hg 2 1 • Place a 2 in front of Hg to balance mercury. • Now there are two mercury atoms on the reactant side and there are two mercury atoms on the product side. • Mercury (Hg) is balanced.

  24.  THE EQUATION IS BALANCED 2 HgO  2 Hg + O2 Element Reactant Side Product Side Hg 2 2 O 2 2

  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. Write and Balance the Equation 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.

  28. The Unbalanced (skeleton) Equation 2 H2SO4(aq) + NaOH(aq) → Na2SO4(aq) + H2O(l) Reactant Side Product Side SO41 1 Na 12 O 1 1 H32 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. 2 2 4 There is one Na on the reactant side and there aretwo Na on the product side. Place a 2 in front of NaOH to balance Na.

  29.  THE EQUATION IS BALANCED  2 2 H2SO4(aq) + NaOH(aq) → Na2SO4(aq) + H2O(l) Reactant Side Product Side SO41 1 Na 2 2 O 21 H42 2 4 There are 4 H on the reactant side and two H on the product side. Place a 2 in front of H2O to balance H.

  30. Write and Balance the Equation When gaseous butane (C4H10) is completely burned in oxygen, the products are carbon dioxide and water.

  31. Balance the Equation 4 C4H10 (g) + O2 (g)→ CO2(g) + H2O(l) Reactant Side Product Side C4 1 H 10 2 O 2 3 4 9 There are four C on the reactant side and there isone C on the product side. Place a 4 in front of CO2 to balance C.

  32. 4 5 C4H10 (g) + O2 (g)→ CO2(g) + H2O(l) Reactant Side Product Side C4 4 H 10 2 O 2 9 10 13 There are 10 H on the reactant side and there aretwo H on the product side. Place a 5 in front of H2O to balance H.

  33. 8 8 20 20 26 10 2 C4H10 (g) + O2 (g)→ 4 CO2(g) + 5 H2O(l) 8 Reactant Side Product Side C4 4 H10 10 O 2 13 There is no whole number coefficient that can be placed in front of O2 to balance O. To balance O double all of the other coefficients.

  34. 10 2 8 C4H10 (g) + O2 (g)→ CO2(g) + H2O(l)  THE EQUATION IS BALANCED  13 Reactant Side Product Side C8 8 H20 20 O 2 26 26 There are now 26 O on the product side. Place a 13 in front of O2 to balance O.

  35. Al + CuCl2 Cu + AlCl3 Al Cu Cl Balancing Example Aluminum and copper(II) chloride react to form copper and aluminum chloride. 2 3 3 2 1 1 1 1 2 3 2   2  6  3 3  6 

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

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

  38. 2Ca(s) + O2(g) 2CaO(s) In a synthesis reaction, there is only one product. 4Al(s) + 3O2(g) 2Al2O3(s) S(s) + O2(g)  SO2(g) 2Al(s) + 3Cl2(g)  2AlCl3(s) Na2O(s) + H2O(l)  2NaOH(aq)

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

  40. In a decomposition reaction, there is only one reactant. 2KClO3(s)  2KCl(s) + O2(g) 2Ag2O(s)  4Ag(s) + O2(g) 2NaNO3(s)  2NaNO2(s) + O2(g) 2H2O2(l)  2H2O(l) + O2(g)

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

  42. Mg(s) + HCl(aq)  H2(g) + MgCl2(aq) Notice that the formula for the magnesium-chloride compound is MgCl2, not MgCl. You will have to be able to write correct formulas based on your understanding of the ions involved; (Mg  Mg2+, therefore 2 Cl- ions are required to balance the charges).

  43. Sodium (Na) will displace any element below it from one of its compounds. The Activity Series An atom of an element in the activity series will displace an atom of any element below it from one of its compounds . MetalsKCaNaMgAlZnFeNiSnPbHCuAgHg increasing activity

  44. Magnesium is above lead in the activity series. Mg(s) + PbS(s)  MgS(s) + Pb(s) MetalsMgAlZnFeNiSnPb

  45. Silver is below copper in the activity series. Ag(s) + CuCl2(s) no reaction ? MetalsPbHCuAgHg

  46. Chlorine is above bromine in the activity series. There is also a halogen activity series – it has the same order as the halogens are found in the Periodic Table Cl2(g) + CaBr2(s)  CaCl2(aq) + Br2(aq) ? HalogensF2Cl2Br2I2

  47. A combines with D Double Displacement Reaction 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. And B combines with C AB+ CD  AD + BC

  48. One or more of the following will accompany a Double Displacement Reaction • formation of a precipitate • formation of a gas (release of bubbles) • release or absorption of heat • formation of a molecular compound (e.g., H2O) 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 acid + base → salt + water HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l) H2SO4(aq) + 2NaOH(aq)  Na2SO(aq) + 2H2O(l)

  50. Formation of an Insoluble Precipitate AgNO3(aq) + NaCl(aq) AgCl(s) + NaNO3(aq) Pb(NO3)2(aq) + 2KI(aq) PbI2(s) + 2KNO3(aq)

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