Stoichiometry

# Stoichiometry

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## Stoichiometry

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##### Presentation Transcript

1. Stoichiometry The Study of Quantitative Relationships

2. 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.

3. 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.

4. 1st Step: Balanced Equation • 2PbS + 3O2 2PbO + 2SO2

5. 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?

6. 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.

7. 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

8. 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.

9. 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

10. 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

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

12. 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.