2. Stoichiometry The Study of Quantitative Relationships

3. 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.

4. 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.

5. 1 st Step: Balanced Equation  2PbS + 3O 2  2PbO + 2SO 2

6. 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?

7. 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.

8. 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

9. 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.

10. 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

11. 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

12. Reality Check  If we worked in industry, we would report the mass of PbO produced not the moles of PbO produced.

13. 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.