Stoichiometry The Study of Quantitative Relationships.

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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  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  2PbS + 3O 2  2PbO + 2SO 2

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 + 3O 2  2PbO + 2SO 2

The BCA Table  Equation: 2PbS + 3O 2  2PbO + 2SO 2 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 Relationships  From the mole ratios between PbS and O 2, we determine we need 0.40 mol of PbS to react 0.60 mol O 2.  2PbS + 3O 2  2PbO + 2SO 2  0.60 mol O 2 x 2 mol PbS = 0.40 mol PbS 3 mol O 2

Completed BCA Table  Equation: 2PbS + 3O 2  2PbO + 2SO 2 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 = g / mol  O (16.00 g / mol ) x 1 = g / mol  g / mol g / mol = g / mol PbO  0.40 mol PbO x g PbO = g PbO 1 mol PbO  g PbO is produced in the reaction.