stoichiometry – chemical quantities notes

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Stoichiometry – Chemical Quantities Notes

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Stoichiometry – Chemical Quantities Notes. Stoichiometry. Stoichiometry – Study of quantitative relationships that can be derived from chemical formulas and chemical equations Mole ─ Mole Relationship need a balanced equation - PowerPoint PPT Presentation

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Page 1: Stoichiometry – Chemical Quantities Notes

Stoichiometry – Chemical Quantities Notes

Page 2: Stoichiometry – Chemical Quantities Notes

StoichiometryStoichiometry – Study of quantitative relationships that can be derived

from chemical formulas and chemical equationsMole ─ Mole Relationship

need a balanced equationMole Ratio – the ratio of moles of one substance to moles of another

substance in a balanced chemical equationThe coefficients in a balanced equation give the relative numbers of

molecules, as well as, the relative number of moles.

CO(g) + 2H2(g) CH3OH(l)

Ex: How many moles of O2 are required to produce 10. moles of CO2?2 CO + O2 2 CO2

1 mol CO = 2 mol H2 = 1 mol CH3OH

10. mol CO2 x __________ mol CO2

mol O2

2

1 = 5.0 mol O2

Page 3: Stoichiometry – Chemical Quantities Notes

What other relationships do we have for the mole?

• 1 mol = 6.022 x 1023 atoms / molecules / particles

• 1 mol = [molar mass] g

We can add these mole relationships on either end of the mole ratio:

# unit A x 1 mol A x mol B x __ unit B = # unit B

_ unit A _ mol A 1 mol B

mole relationship mole ratio mole relationship

(switch units) (switch substances) (switch units)

*A is the GIVEN substance & B is the WANTED substance

Page 4: Stoichiometry – Chemical Quantities Notes

Mass A – Mole B

Ex: Calculate moles of O2 produced if 2.50 g KClO3 decomposes completely:

2 KClO3 2 KCl + 3 O2

x ______________mol O2

mol KClO3

3

2

x ________________mol KClO3

122.6

1

g KClO3

2.50 g KClO3 =

0.0306 mol O2

K 1 x 39.1 = 39.1

Cl 1 x 35.5 = 35.5

O 3 x 16.0 = 48.0

122.6 g/mol

Page 5: Stoichiometry – Chemical Quantities Notes

Mass A – Mass B

Ex: Determine the mass of NaCl that decomposes to yield 355 g Cl2

2NaCl 2 Na + Cl2355 g Cl2

mol Cl2

g Cl271.0

1 x _____________ x ______________ mol Cl2

mol NaCl

1

2 x ______________ mol NaCl

g NaCl58.5

1=

585 g NaCl

Cl 2 x 35.5 = 71.0 g/mol

Na 1 x 23.0 = 23.0

Cl 1 x 35.5 = 35.5

58.5 g/mol

Page 6: Stoichiometry – Chemical Quantities Notes

Mole A – Mass BEx: Calculate the number of grams of oxygen

required to react exactly with 4.30 mol of propane, C3H8, in the reaction by the following balanced equation:

C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g)

4.30 mol C3H8 mol O2

mol C3H81

5 x _____________ x __________ mol O2

g O2

1

32.0=

688 g O2O 2 x 16.0 = 32.0 g/mol