Stoichiometry The Study of Quantitative Relationships.
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Transcript of Stoichiometry The Study of Quantitative Relationships.
Stoichiometry
The Study of Quantitative Relationships
What is Stoichiometry?
Stoichiometry is the study of
quantitative relationships between
the amounts of reactants used and
the amounts of products produced in
a chemical reaction. Stoichiometry is
based on the law of conservation of
mass.
Using Stoichiometry
Start with a balanced equation for the chemical reaction!
Lead (II) sulfide reacts with oxygen
gas to produce lead (II) oxide and
sulfur dioxide.
1st Step: Balanced Equation
2PbS + 3O2 2PbO + 2SO2
Analyzing the Problem
QUESTION: If 0.60 mole of
oxygen were consumed during a
chemical reaction between
oxygen and lead II sulfide how
many GRAMS of lead (II) oxide
would be produced?
Analyzing the Problem
PROBLEM: Determine the mass of
one of the products when the moles
of one reactant in a chemical
reaction is known.
Use a BCA table to make this
calculation easier.
Using Stoichiometry
Start with the balanced equation for the reaction!
Solid lead (II) sulfide reacts with oxygen gas to produce solid lead (II) oxide and sulfur dioxide gas.
2PbS + 3O2 2PbO + 2SO2
The BCA TableEquation: 2PbS + 3O2 2PbO + 2SO2
Before: ? mol .60 mol 0 mol 0 mol
Change - ? mol -.60 mol +__mol __mol
_________________________________________________After 0 mol 0 mol ? mol ? mol
The only information we are given is the amount of oxygen consumed.
Mole RelationshipsFrom the mole ratios between PbS and O2, we determine we need 0.40 mol of PbS to react 0.60 mol O2.
2PbS + 3O2 2PbO + 2SO2
0.60 mol O2 x 2 mol PbS = 0.40 mol PbS
3 mol O2
Completed BCA Table
Equation: 2PbS + 3O2 2PbO + 2SO2
Before: .40 mol .60 mol 0 mol 0 mol
Change - .40 mol - .60 mol +.40 mol +.40 mol ___________________________________________After 0 mol 0 mol .40 mol .40 mol
Reality Check
If we worked in industry, we would
report the mass of PbO produced not
the moles of PbO produced.
What Mass of PbO Was Produced?
Using the molar mass of PbO convert 0.40 moles of PbO to grams of PbO.
Pb (207.2 g/mol) x 1 = 207.2 g/mol
O (16.00 g/mol) x 1 = 16.00 g/mol
207.2 g/mol + 16.00 g/mol = 223.2 g/mol PbO
0.40 mol PbO x 223.2 g PbO = 89.28 g PbO
1 mol PbO
89.28 g PbO is produced in the reaction.