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

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### 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.

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

- 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

- 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.

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