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Mr. Ramos.   The gasoline in an automobile gas tank has a mass of 60.0 kg and a density of 0.752 g/cm 3. What is its volume in cm 3 ?

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Presentation on theme: "Mr. Ramos.   The gasoline in an automobile gas tank has a mass of 60.0 kg and a density of 0.752 g/cm 3. What is its volume in cm 3 ?"— Presentation transcript:

1 Mr. Ramos

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4  The gasoline in an automobile gas tank has a mass of 60.0 kg and a density of 0.752 g/cm 3. What is its volume in cm 3 ?

5  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  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 10 23, then the mass of that element is changed from amu to grams.  1 mole = 6.022 x 10 23 (Avogadro’s Number)

7  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 10 23 particles

8  1 mole of M&M’s = 6.022 x 10 23 M&M’s ◦ That’s a lot of M&M’s (Yummy)  1 mole of U.S. dollars = 6.022 x 10 23 U.S. dollars ◦ I would be rich! YES!  1 mole of carbon = 6.022 x 10 23 carbon atoms = 12.01 grams of carbon

9  6CO 2 + 6H 2 O + Sunlight  C 6 H 12 O 6 + 6O 2  The formula above represents the photosynthesis equation.  How many grams of glucose are produced when 12 moles of carbon dioxide are consumed?

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

11  Water is formed when hydrogen gas reacts explosively with oxygen as according to the following balanced equation. ◦ O 2 (g) + 2H 2 (g)  2H 2 O (g)  How many moles of H 2 O result from the complete reaction of 24.6 mol of O 2 ? Assume that there is more than enough H 2.

12  How many moles of ammonia (NH 3 ) are produced when 0.60 mol of nitrogen (N 2 ) reacts with Hydrogen (H 2 )? ◦ N 2 (g) + 3H 2 (g)  2NH 3 (g)

13  For any closed system, mass cannot be created or destroyed. ◦ Mass can only be transferred. Reactants Products

14  Chemical equations must always be balanced.  6CO 2 + 6H 2 O  C 6 H 12 O 6 + 6O 2 Carbon = 6 Oxygen = 18 Hydrogen = 12 Carbon = 6 Oxygen = 18 Hydrogen = 12 Reactants Products

15  1. Write a skeletal equation by writing chemical formulas for each of the reactants and products.  SiO 2 (s) + C (s)  SiC (s) + CO (g)

16  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) ◦ SiO 2 (s) + C (s)  SiC (s) + 2CO (g)

17  SiO 2 (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) ◦ SiO 2 (s) + 3C (s)  SiC (s) + 2CO (g)

18  SiO 2 (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  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.  SiO 2 (s) + 3C (s)  SiC (s) + 2CO (g) Silicon = 1 Oxygen = 2 Carbon = 3 Silicon = 1 Oxygen = 2 Carbon = 3

20  C 8 H 18 (l) + O 2 (g)  CO 2 (g) + H 2 O (g)

21  Fe (s) + 3HCl (aq)  FeCl 3 (aq) + H 2 (g)


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