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Chapter 9 is Stoichiometry!!

I. Dimensional Analysis. Chapter 9 is Stoichiometry!!. I can help you count moles!!!. I. Dimensional Analysis. The Art of Counting Without Counting. Composition stoichiometry , which you studied in chapter three, involves mass relationships of elements within compounds .

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Chapter 9 is Stoichiometry!!

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  1. I Dimensional Analysis Chapter 9is Stoichiometry!! I can help you count moles!!!

  2. I Dimensional Analysis The Art of Counting Without Counting Composition stoichiometry, which you studied in chapter three, involves mass relationships of elements within compounds. Reaction stoichiometryinvolves the mass relationships between reactants and products in a chemical reaction. In other words, you are converting amounts of one substance to amounts of another substance in a reaction. **All reaction stoichiometry questions start with a BALANCED CHEMICAL EQUATION!!!!

  3. I Dimensional Analysis Reaction stoichiometry problems Easy as 1,2,3,4 1. Write the formula equation and balance it. 2. Convert grams of compound A to moles of compound A using the molar mass of compound A. 3. Use balanced equation to find the mole ratio (this is where you make the switch from compound A to compound B) 4. Convert moles of compound B to moles of compound B using the molar mass of compound B.

  4. I Dimensional Analysis Conversions you will need to be able to understand and use: How many atoms (molecules or particles or chairs or Dalmatians) are in a mole? 6.02 X 1023 How do you find how many grams are in a mole of a substance? Look at your periodic table!! Add up the masses of the elements in the compound! How many liters of a gas are in a mole? (this one is new) 1 mole of ANY gas = 22.4 Liters What is a mole ratio? (this one’s new, too) You must have a correctly balanced equation for this to be correct. ZnCl2 + 3O2 → Zn(ClO3)2 1 mole of ZnCl2 3 mole O2 1 mole of ZnCl2 1 mole Zn(ClO3)2 3 moles of O2 1 mole Zn(ClO3)2

  5. Stoichiometry Roadmap liters of A (gases only) • liters of B • (gases only) X 22.4 L / 22.4 L X 22.4 L / 22.4 L / molar mass / molar mass Mole ratio grams of A moles of A moles of B grams of B X molar mass X molar mass X Avog. # / Avog. # / Avog. # X Avog. # particles of A particles of B

  6. Example 1 Tin (II) fluoride, SnF2, is used in some toothpastes. It is made by the reaction of tin with hydrogen fluoride according to the following equation: Sn(s) + 2HF → SnF2(s) + H2(g) How many grams of SnF2 are produced from the reaction of 30.00 g of HF with Sn?

  7. I Dimensional Analysis How many grams of SnF2 are produced from the reaction of 30.00 g of HF with Sn? Sn(s) + 2HF → SnF2(s) + H2(g) Check the road map Grams A → mole A →mole ratio →moleB → grams B 30.00 g HF 1 mol HF 1 mol SnF2 156.71 g SnF2 20.01 g HF 2 mol HF 1 mol SnF2 = 117.5 g SnF2 SIG FIGS COUNT!!

  8. Example 2 Laughing gas (nitrous oxide) is sometimes used as an anesthetic in dentistry. It is produced when ammonium nitrate is decomposed according to the following reaction: NH4NO3 (s) →N2O (g) + 2H20 (l) How many moles of NH4NO3 are required to produce 33.0 g of N2O?

  9. I Dimensional Analysis NH4NO3 (s) →N2O (g) + 2H20 (l) How many moles of NH4NO3 are required to produce 33.0 g of N2O? Check the road map: Grams X → mol X → mole ratio → mol Y 33.0 g N2O 1 mol N2O 1 mol NH4NO3 = .750 mol NH4NO3 44.02 g N2O 1 mol N2O

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