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.