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Law of Conservation of Matter
Mass-Mass Problems
The word stoichiometry derives from two Greek words:
stoicheion (meaning "element") and metron
(meaning "measure"). Stoichiometry deals with
calculations about the masses and volumes of
reactants and products involved in a chemical reaction.
Every chemical reaction has it's characteristic
proportions. The method of obtaining these proportions
from chemical formulas, equations, atomic weights and
molecular weights, and the determination of what and
how much is used and produced in chemical processes,
is the major emphasis of Stoichiometry.
Proportional Relationships
• Stoichiometry
– mass relationships between substances
in a chemical reaction
– based on the mole ratio
• Mole Ratio
– indicated by coefficients in a balanced
equation
2 Mg + O2  2 MgO
What is the mole ratio of magnesium metal to magnesium oxide?
What is the mole ratio of magnesium oxide to oxygen?
2:1
2:2
Steps to follow to solve a stoichiometry mass-mass problem
1. Write out and balance the chemical reaction
2. Determine what is the given and unknown compounds
3. Convert given mass to given moles by dividing the molar
mass.
4. Use the molar ratio from the balanced equation (unknown
over known) to convert to unknown moles.
5. Convert unknown moles to unknown mass by multiplying
molar mass.
Molar Ratio
Given Mass (g)
1 (mol) of
known
unknown mol
Molar Mass of
known (g)
known mol
Molar Mass of
unknown (g)
1 (mol) of
unknown
Stoichiometry Problems
When N2 combines with H2 they produce 5.63 g of
NH3. How many grams of N2 were needed to
produce the ammonia gas?
N2
4.63g
_____?
5.63 g NH3
+
3 H2  2 NH3
1.00g
1 mol NH3 1 mol N2
17.04 g NH3 2 mol NH3
5.63 g
28.02 g N2
1 mol N2
=
(5.63 x 1 x 1 x 28.02) ÷ (17.04 x 2 x 1) = 4.63 g N2
How many grams of H2 are needed for
the reaction to take place? Remember
the law of conservation of matter.
Stoichiometry Problems
6.50 grams of aluminum reacts with an excess of
oxygen. How many grams of aluminum oxide are
formed?
4 Al
6.50 g
6.50 g Al
1 mol Al
26.98 g Al
+
3 O2  2 Al2O3
5.78g
12.28g
_____?
2 mol Al2O3 101.96 g Al2O3
4 mol Al
1 mol Al2O3
=
(6.50 x 1 x 2 x 101.96) ÷ (26.98 x 4 x 1) = 12.28 g Al2O3
How much oxygen is needed for the reaction to take
place?
Stoichiometry Problems
Acetylene gas C2H2 burns in a combustion reaction for
welding. How many grams of C2H2 are burned if the
reaction produces 75.0 g of CO2?
2 C2H2
+ 5 O2 
22.19g
_____?
75.0 g CO2
1 mol CO2
4 CO2
+
2 H2O
75.0 g
2 mol C2H2
44.01 g CO2 4 mol CO2
26.04 g C2H2
1 mol C2H2
=
(75.0 x 1 x 2 x 26.04) ÷ (44.01 x 4 x 1) = 22.19 g C2H2