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Conversions & Balancing Equations

Conversions & Balancing Equations. Mr. Ramos. What are Conversions?. Conversion means to change something “appearance.” Example: 1 foot = 12 inches Example: 1 meter = 100 centimeters 1ft = 12in is known as a conversion factor

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Conversions & Balancing Equations

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  1. Conversions & Balancing Equations Mr. Ramos

  2. What are Conversions? • Conversion means to change something “appearance.” • Example: 1 foot = 12 inches • Example: 1 meter = 100 centimeters • 1ft = 12in is known as a conversion factor Cancel the units of the “Given” with the units of the conversion factor’s denominator.

  3. Conversion Examples

  4. Chemistry Example • The gasoline in an automobile gas tank has a mass of 60.0 kg and a density of 0.752 g/cm3. What is its volume in cm3?

  5. Where are the Units? • What is the mass of Hydrogen, Carbon, or Oxygen? • Go to the periodic table and try to find the corresponding unit associated with the mass of these elements. • Hydrogen’s mass is 1.01 what? • Carbon’s mass is 12.01 what? • Oxygen’s mass is 16.00 what? • Where are the UNITS?

  6. Where are the Units? • The atomic mass unit (amu), or Dalton, is used to indicate the mass of atoms and molecules. • Hydrogen has a mass of 1.01 amu • Carbon has a mass of 12.01 amu • Oxygen has a mass of 16.00 amu • If, however, we have a fixed amount of any element that equals 6.022 x 1023, then the mass of that element is changed from amu to grams. • 1 mole = 6.022 x 1023 (Avogadro’s Number)

  7. The Mole • The element hydrogen has a mass of 1.01 amu. • However, 1 mole of hydrogen has a mass of 1.01 grams. • The element carbon has a mass of 12.01 amu. • However, 1 mole of carbon has a mass of 12.01 grams. • The element oxygen has a mass of 16.00 amu. • However, 1 mole of oxygen has a mass of 16.00 grams. • Do you notice the pattern? • 1 mole of anything contains 6.022 x 1023 particles

  8. The Mole: Avogadro’s # • 1 mole of M&M’s = 6.022 x 1023 M&M’s • That’s a lot of M&M’s (Yummy) • 1 mole of U.S. dollars = 6.022 x 1023 U.S. dollars • I would be rich! YES! • 1 mole of carbon = 6.022 x 1023 carbon atoms = 12.01 grams of carbon

  9. Conversions Involving Moles • 6CO2 + 6H2O + Sunlight  C6H12O6 + 6O2 • The formula above represents the photosynthesis equation. • How many grams of glucose are produced when 12 moles of carbon dioxide are consumed?

  10. Solve by Converting • Sodium chloride, NaCl, forms by the following reaction between sodium and chlorine. • 2Na (s) + Cl2 (g)  2NaCl (s) • How many moles of NaCl result from the complete reaction of 3.4 mol of Cl2? Assume that there is more than enough Na.

  11. Solve by Converting • Water is formed when hydrogen gas reacts explosively with oxygen as according to the following balanced equation. • O2 (g) + 2H2 (g)  2H2O (g) • How many moles of H2O result from the complete reaction of 24.6 mol of O2? Assume that there is more than enough H2.

  12. Solve by Converting • How many moles of ammonia (NH3) are produced when 0.60 mol of nitrogen (N2) reacts with Hydrogen (H2)? • N2 (g) + 3H2 (g)  2NH3 (g)

  13. Law of Conservation of Mass • For any closed system, mass cannot be created or destroyed. • Mass can only be transferred. Reactants Products

  14. Balanced Chemical Equations • Chemical equations must always be balanced. • 6CO2 + 6H2O  C6H12O6 + 6O2 Carbon = 6 Oxygen = 18 Hydrogen = 12 Carbon = 6 Oxygen = 18 Hydrogen = 12 Products Reactants

  15. Writing Balanced Chemical Equations • 1. Write a skeletal equation by writing chemical formulas for each of the reactants and products. • SiO2 (s) + C (s)  SiC (s) + CO (g)

  16. Writing Balanced Chemical Equations • SiO2 (s) + C (s)  SiC (s) + CO (g) • 2. If an element occurs in only one compound on both sides of the equation, balance it first. If there is more than one such element, balance metals before nonmetals. • Begin with Si: 1 Si atom  1 Si atom • Si is already balanced • Balance O next: 2 O atoms  1 O atom • To balance O, put a 2 before CO (g) • SiO2 (s) + C (s)  SiC (s) + 2CO (g)

  17. Writing Balanced Chemical Equations • SiO2 (s) + C (s)  SiC (s) + 2CO (g) • 3. If an element occurs as a free element on either side of the chemical equation, balance it last. Always balance free elements by adjusting the coefficient on the free element. • Balance C: 1 C  3 C • To balance C, put a 3 before C (s) • SiO2 (s) + 3C (s)  SiC (s) + 2CO (g)

  18. Writing Balanced Chemical Equations • SiO2 (s) + 3C (s)  SiC (s) + 2CO (g) • 4. If the balanced equation contains coefficient fractions, clear these by multiplying the entire equation by the appropriate factor. • This step is not necessary in this example.

  19. Writing Balanced Chemical Equations • 5. Check to make certain the equation is balanced by summing the total number of each type of atom on both sides of the equation. • SiO2 (s) + 3C (s)  SiC (s) + 2CO (g) Silicon = 1 Oxygen = 2 Carbon = 3 Silicon = 1 Oxygen = 2 Carbon = 3

  20. Writing Balanced Chemical Equations: Solve • C8H18 (l) + O2 (g)  CO2 (g) + H2O (g)

  21. Writing Balanced Chemical Equations: Solve • Fe (s) + 3HCl (aq)  FeCl3 (aq) + H2 (g)

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