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CHEMISTRY

CHEMISTRY. Mass Relations In Chemical Reactions. What Does a Chemical Equation Mean. A Chemical equation represents the substances that undergo a chemical change and the products that are formed. This is done in such a way to illustrate the Law of Conservation of Matter. EXAMPLE:

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CHEMISTRY

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  1. CHEMISTRY Mass Relations In Chemical Reactions

  2. What Does a Chemical Equation Mean • A Chemical equation represents the substances that undergo a chemical change and the products that are formed. This is done in such a way to illustrate the Law of Conservation of Matter. • EXAMPLE: • Nitrogen reacts with hydrogen to produce ammonia • N2 + H2 ------ > NH3 1 3 2 This is a balanced equation 1 moles of nitrogen 3 moles of hydrogen 2 moles of ammonia

  3. The Mole Tree • Of course we mean the Chemical mole 1.00 mol = molar mass = molar volume = molar numbers (sum of at. Mass) ( 22.4 l @STP) (6.02 x 1023 p)

  4. Mole Relationship • Vol1 Vol2 • Grams1 Mole1 Mole2 Grams2 • #’s1 #’s2 If we go from grams of one substance to grams of another, we must go through the mole. If we go from grams of one substance to #’s of another, we go through the mole.

  5. The Mole Ratio • The ratio of moles in a balanced chemical equation is set by the law of conservation. • N2 + 3 H2 ----- > 2 NH3 • The mole ratio is 1 : 3 : 2 • The only way to convert from mass, volume or numbers of one substance to another is through the mole ratio.

  6. Example • N2 + 3 H2 ----- > 2 NH3 • How many moles of ammonia are produced if 10.0 g of hydrogen reacting with an excess of nitrogen? • Solution: • Use the map – The mole tree • Vol1 Vol2 • | | • Grams1 -- Mole1 ----- Mole2 -- Grams2 • | | • #’s1 #’s2 10.0 g H2 mol H2 mol NH3 Find Given

  7. The O’Lamar Factor • Use the unit factor method • 10.0 g H2 mol H2 mol NH3 What is the mole ratio and where is it found? 2 1.00 =3.03 mol NH3 3 2.02 g H2 mol H2 Grams and moles means? Molar mass N2 + 3 H2 ----- > 2 NH3

  8. And Again • Problem: • How many grams of water are produced in the combustion of 254 g of heptane ( C7H16)? • Solution: Write and balance the equation. • C7H16 + O2 --- CO2 + H2O • Use the tree : grams-moles-moles-grams • 254 g C7H16 x mol C7H16 x mol H2O x g H2O 11 7 8 1.00 8 18.0 1 mol C7H16 mol H2O 100 g C7H16 1.00 366 g H2O

  9. Limiting Reactant • The limiting reactant in a chemical reaction is the reactant used up first. • Example: When a match is burning and goes out its because the fuel and wood in the match are consumed and not the oxygen in the room. The limiting reactant is the match and the oxygen is in excess.

  10. Limiting Reactant A student mixed 25.0 g of Sulfur with 25.0 g of Oxygen. • How many grams of sulfur trioxide are formed? • How many grams of excess reactant are left? Write and balance the equation In order to solve the problem we must find the limiting reactant ( which one is consumed first ) 25g 25g 2 S + O2 ---> SO3 3 2

  11. Limiting Reactant Pick one reactant and solve for the mass of the other. 25.O g S x 1.00 mol S x 3 mol O2 x 32.0 g O2 32.0g S 2 mol S 1.00 mol O2 What this tells us is that in order to react ALL 25.g of sulfur we need 37.6 g of oxygen. Since there are only 25.0 g the oxygen is used first and is the LR 25g 25g 2 S + O2 ---> SO3 3 2 = 37.6 g O2

  12. Which One is It • Use the LR to find the mass of the product. 25.O g O2 x mol O2 x mol SO3 x g SO3 How much excess Sulfur is not reacted? Start with the LR 25.0 g O2 1.00 mol O2 x 2 mol S x 32.1 g S 32.0 g O2 3 mol O2 1.00 mol S 16.7 g S used 25.0 g- 16.7 g = 8.30 g XS 1.00 2 80.1 3 mol O2 1.00 mol SO3 32.0 g O2 = 41.7 g SO3 16.7 g S

  13. What about Volume of a Gas At STP? • 1.00 mole = 22.4 liters at Standard temperature and standard pressure. • Standard Temperature = 0.000C or 273 K • Standard Pressure = 760 mm Hg = 1.00 atm = 101.3 kpa • If we wanted to change grams of one substance to volume in liters to another we would follow the map. • Vol1Vol2 • | | • Grams1 --- Mole1 ----- Mole2 --- Grams2 • | | • #’s1 #’s2

  14. Problem • What volume of octane at STP is needed to produce 2.45 Tons of Carbon dioxide during the combustion of Octane ( C8H18 )? • First the Reaction: C8H18 + O2 --- > CO2 + H2O First a little conversion 2.45 T x 2.0 x 103 lbs x 454 g = 1.00 T 1.00 lbs 2 25 16 18 2.22 x 106 g CO2

  15. Now Solve • 2C8H18 + 25 O2 --- > 16 CO2 + 18 H2O 2.22 x 106 g CO2 x mol CO2 1.00 x mol C8H18 2 x L C8H18 22.4 44.0 g CO2 16 mol CO2 1.00 mol C8H18 1.41 x 105 L C8H18

  16. It’s All About The Numbers • Remember that 1.00 moles = 6.02 x 1023 particles • How many molecules of hydrogen are released when 1.00 g of Zinc is dissolved in hydrochloric acid? • Zn + 2 HCl ---- > ZnCl2 + H2 1.00 g Zn x 1.00 mol Zn x 1.00 mol H2 x 6.02 x 1023molec.H2 65.4 g Zn 1.00 mol Zn 1.00 mol H2 9.20 x 1021molec. H2

  17. Repeat After Me • Grams, moles,moles,grams • Grams, moles,moles,liters • Grams, moles,moles,numbers • Grams, moles,moles,grams • Grams, moles,moles,liters • Grams, moles,moles,numbers • Liters, moles,moles,grams • Numbers, moles,moles,grams • Grams, moles,moles,grams • Grams, moles,moles,grams

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