Transcript Stoichiometry

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
Mole Ratio – the ratio of moles of one substance to moles of another
substance in a balanced chemical equation
The 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)
1 mol CO = 2 mol H2 = 1 mol CH3OH
Ex: How many moles of O2 are required to produce 10. moles of CO2?
2 CO + O2  2 CO2
10. mol CO2 x __________
1 mol O2
= 5.0 mol O2
2 mol CO2
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
(switch units)
(switch substances)
mole relationship
(switch units)
*A is the GIVEN substance & B is the WANTED substance
Mass A – Mole B
Ex: Calculate moles of O2 produced if 2.50 g
KClO3 decomposes completely:
2 KClO3  2 KCl + 3 O2
2.50 g KClO3 x ________________
1 mol KClO3
122.6 g KClO3
3 mol O2
x ______________
=
2 mol KClO3
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
0.0306 mol O2
Mass A – Mass B
Ex: Determine the mass of NaCl that
decomposes to yield 355 g Cl2
2NaCl  2 Na + Cl2
355 g Cl2
1 mol Cl2
2 mol NaCl
58.5 g NaCl
x _____________
x ______________
x ______________
=
71.0 g Cl2
1 mol Cl2
1 mol NaCl
Cl 2 x 35.5 = 71.0 g/mol
585 g NaCl
Na 1 x 23.0 = 23.0
Cl 1 x 35.5 = 35.5
58.5 g/mol
Mole A – Mass B
Ex: 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)
5 mol O2
32.0 g O2
4.30 mol C3H8 x _____________
x __________
1 mol C3H8
1 mol O2
O 2 x 16.0 = 32.0 g/mol
688 g O2
=