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Ch. 9: Stoichiometry Stoichiometry – pronounced stoy-key-ah-meh-tree. A balanced chemical equation can be interpreted on a molecular level, a mole level,

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Presentation on theme: "Ch. 9: Stoichiometry Stoichiometry – pronounced stoy-key-ah-meh-tree. A balanced chemical equation can be interpreted on a molecular level, a mole level,"— Presentation transcript:

1 Ch. 9: Stoichiometry Stoichiometry – pronounced stoy-key-ah-meh-tree. A balanced chemical equation can be interpreted on a molecular level, a mole level, or a mass level. Ex) N2 + 3 H2  2 NH3 The relationship between N 2 and H 2 is: 1 molecule of N 2 = 3 molecules H 2 Or:

2 Stoichiometry We can also establish similar relationships between the product and the reactants. This allows us to convert any amount of N2, H2, and NH3 into a quantity of the others. Analogy: Making a bicycle. 1 frame + 2 wheels  1 bicycle What if you had 40 frames? What if you made 75 bicycles?

3 Stoichiometry Fill in the table below. Amount of N2 Amount of H2 Amount of NH3 1 molecule3 molecules2 molecules 18 molecules 120 molecules 5,400 molecules

4 Stoichiometry It is impractical to do most reactions in terms of molecules. The coefficients can be viewed as “moles.” Thus, 1 mole of N2 plus 3 moles of H2 will produce 2 moles of NH3.

5 Stoichiometry Fill in the table below. Amount of N2 Amount of H2 Amount of NH3 2.4 moles moles 6.4 moles 27.3 moles

6 Stoichiometry Finally, we can calculate any mass from a mole amount using the molar mass of each compound. Amount of N2 Amount of H2 Amount of NH3 1 mole3 moles2 moles moles0.12 moles0.080 moles

7 Stoichiometry Converting a mass-to-mass of any reactant or product involves three steps. Mass of AMass of B Moles of A Divide by Molar Mass of A Moles of B Multiply by the ratio of B / A Multiply by Molar Mass of B

8 Stoichiometry What mass, in grams, can be produced by the complete reaction of 3.40grams of N2?

9 Stoichiometry What mass of PH3 is possible from the complete reaction of 150.grams of Ca3P2? The balanced reaction is: Ca3P2 + 6 H2O  3 Ca(OH)2 + 2 PH3 What three pieces of information will we need?

10 Stoichiometry What mass of Br2 is required to react with 0.185grams of Al? The balanced reaction is: 2 Al + 3 Br2  2 AlBr3

11 Stoichiometry Limiting reactant – amounts used in a reaction may have one (or more) reactant in excess and one in a lesser amount – this is the limiting reactant. Ex) Back to our bicycle assembly – what if the warehouse included 200 frames and 350 wheels. How many bicycles could you make? What is the limiting “reactant”?

12 Stoichiometry Mole-amounts require the application of the mole-to-mole amounts. Using the reaction: 6 Na + N2  2 Na3N Amount of Na Amount of N2 Limiting?Amount of Na3N 18 moles6.0 moles 1.5 moles0.22 moles 0.54 moles0.12 moles 3.6 moles0.55 moles

13 Stoichiometry Mass amounts require application of the three-step process. In the reaction below, 4.55g of Fe2O3 and 2.85g of CO are allowed to react. What mass of Fe metal is possible? Fe2O3 + 3 CO  2 Fe + 3 CO2

14 Stoichiometry The amount of product possible based on the limiting reactant is sometimes called the theoretical yield. The percent yield is then the actual yield divided by the theoretical yield times 100.

15 Stoichiometry When 100.g of CS2 reacts with 100.g of Cl2, 65.0g of CCl4 are produced. Calculate the percent yield of this reaction.


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