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

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Stoichiometry

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Stoichiometry

Stoichiometry

The Study of Quantitative Relationships


What is stoichiometry

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

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.


1 st step balanced equation

1st Step: Balanced Equation

  • 2PbS + 3O2 2PbO + 2SO2


Analyzing the problem

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 problem1

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 stoichiometry1

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 table

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

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

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

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

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.


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