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Chapter 9: Stoichiometry. A. Introduction to Stoichiometry: Stoichiometry – the calculation of an unknown quantity of a substance from a known value of a different substance in a chemical reaction

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chapter 9 stoichiometry
Chapter 9: Stoichiometry

A. Introduction to Stoichiometry:

  • Stoichiometry – the calculation of an unknown quantity of a substance from a known value of a different substance in a chemical reaction

- like mole conversions, except the coefficients in the rxn. matter

- coefficients = # of moles

- 1 mole = molar mass

= 6.02 x 1023 particles

= 22.4 L

slide2
Stoichiometry is like a recipe:
    • Reactants = Ingredients
    • Products = What the recipe makes
    • Limiting reagent – reactant that determines how much product can be made (used up first)
    • Excess reagent(s) – reactant(s) not all used up

ex. Brownies (makes 24)

2 cups of mix

3 eggs

¼ cup oil

slide3
Stoichiometry is like a recipe (cont.):

ex. 2 mix + 3 eggs + ¼ oil  24 brownies

If you had 2x the ingredients, how many brownies?

48 brownies

If you had 96 brownies, how many eggs?

12 eggs

If you have 4 cups mix, 4 eggs, ½ cup oil, what is the limiting reagent?

the eggs (need 6, only have 4)

slide4
B. Mole-Mole Stoichiometry:
  • Balance equation (if not done)
  • Start with given (in moles, in this case)
  • Multiply by mole ratio (coefficients from rxn.)
  • Include all units and substance in your work!

ex. 4Al + 3O22Al2O3

How many moles of aluminum oxide can be produced from 6 moles of oxygen?

6 moles O2 x 2 moles Al2O3 =

3 moles O2

4 moles Al2O3

slide5
ex. 6NaOH + Fe2(SO4)3 3Na2SO4 + 2Fe(OH)3

How many moles of Fe(OH)3 are produced from reacting 8 moles of NaOH?

How many moles of Fe2(SO4)3 are reacted to form 1.7 moles of Na2SO4?

8 moles NaOH x 2 moles Fe(OH)3 = 2.67 moles of

6 moles NaOH Fe(OH)3

1.7 moles Na2SO4 x 1 mole Fe2(SO4)3 = 0.57 moles of

3 moles Na2SO4 Fe2(SO4)3

slide6
C. Volume-Volume Stoichiometry:
  • 1 mole = 22.4 L of any gas at STP
  • Start with given (in L, in this case)
  • Convert given (L) to moles
  • Multiply by mole ratio (converts moles of given to moles of find – use coefficients from rxn.)
  • Convert moles of find back to L
  • Cancel out same units, substances, and numbers
  • Solve problem
slide7
ex. 2SO2 + O2 2SO3

How many liters of oxygen are needed to produce 19.8 L of sulfur trioxide?

19.8 L SO3 x 1 mol SO3 x 1 mol O2 x 22.4 L O2

22.4 L SO3 2 mol SO3 1 mol O2

19.8 L SO3 x 1 mol O2

2 mol SO3

Answer:9.9 L O2

slide8
ex. 2NO + O2 2NO2

How many liters of nitrogen monoxide are needed to react with 5.6 L of oxygen to produce nitrogen dioxide?

5.6 L O2 x 1 mol O2 x 2 mol NO x 22.4 L NO

22.4 L O2 1 mol O2 1 mol NO

5.6 L O2 x 2 mol NO

1 mol O2

Answer:11.2 L NO

slide9
D. Mass-Mass Stoichiometry:
  • 1 mole = molar mass (use masses from the P.T.)
  • Start with given (in grams, in this case)
  • Convert given (grams) to moles (calculate the molar mass)
  • Multiply by mole ratio
  • Convert moles of find back to grams
  • Cancel out same units and substances
  • Solve problem
slide10
ex. 2SO2 + O2 2SO3

How many grams of oxygen are needed to produce 19.8 g of sulfur trioxide?

19.8 g SO3 x 1 mol SO3 x 1 mol O2 x 32.0 g O2

80.1 g SO3 2 mol SO3 1 mol O2

SO3 = 80.1 g/mole

O2 = 32.0 g/mole

Answer:3.96 g O2

slide11
ex. 2NO + O2 2NO2

How many grams of nitrogen monoxide are needed to react with 5.6 g of oxygen to produce nitrogen dioxide?

5.6 g O2 x 1 mol O2 x 2 mol NO x 30.0 g NO

32.0 g O2 1 mol O2 1 mol NO

Answer:10.5 g NO

slide12
E. Mixed Stoichiometry Problems:
  • Uses moles, liters, grams and/or molecules
  • Start with given
  • Convert given to moles (unless given is in moles)
  • Multiply by mole ratio
  • Convert moles of find to unit of find in problem (unless find is in moles)
  • Cancel out same units and substances
  • Solve problem
slide13
ex. 2NO + O2 2NO2

How many liters of nitrogen monoxide are needed to react with 8.2 g of oxygen to produce nitrogen dioxide?

8.2 g O2 x 1 mol O2 x 2 mol NO x 22.4 L NO

32.0 g O2 1 mol O2 1 mol NO

Answer:11.48 L NO