Stoichiometry

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# Stoichiometry - PowerPoint PPT Presentation

Stoichiometry. The Study of Quantitative Relationships. What is Stoichiometry?.

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Presentation Transcript

### 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
• 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 GRAMSof 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
• Solid lead (II) sulfide reacts with oxygen gas to produce solid lead (II) oxide and sulfur dioxide gas.
• 2PbS + 3O2 2PbO + 2SO2
The BCA Table
• Equation: 2PbS + 3O2 2PbO + 2SO2Before: ? mol.60 mol0 mol0 molChange - ?mol -.60 mol+__mol__mol
• _________________________________________________After 0mol0mol?mol?mol
• The only information we are given is the amount of oxygen consumed.
Mole Relationships
• From the mole ratios between PbS and O2, we determine we need 0.40 mol of PbSto react 0.60 mol O2.
• 2PbS + 3O2 2PbO + 2SO2
• 0.60 mol O2x 2 molPbS=0.40 molPbS

3 mol O2

Completed BCA Table
• Equation: 2PbS + 3O2 2PbO + 2SO2Before: .40 mol.60 mol0 mol0 molChange -.40mol- .60 mol+.40mol+.40 mol___________________________________________After 0mol0mol.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 PbOWas 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/molPbO
• 0.40 molPbOx 223.2 g PbO= 89.28 g PbO

1 molPbO

• 89.28 g PbOis produced in the reaction.