# Stoichiometry - PowerPoint PPT Presentation

Stoichiometry

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Stoichiometry

## Stoichiometry

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##### Presentation Transcript

2. Adjusting To Reality • This is not the entirestory. In reality, you neverhave the exactamounts of both reactants you need. At the endof the reaction, one reactant has been completely consumed and there is some “left over” of the other reactant. • Let’s take a second look at the reaction between PbS and O2.

3. 2PbS + 3O2 2PbO + 2SO2 • Since oxygen is not costly (free in our atmosphere), it’s usually the reactant there is plenty of and only a certain amount of lead II sulfidewould have been purchased for this reaction. • What would our BCA table look like if we had 0.40 moles of lead (II) sulfide reacting with an abundance (excess) of oxygen?

4. Adjusting To Reality • Equation: 2PbS + 3O2 2PbO + 2SO2Before: .40 mol xs mol 0 mol 0 mol Change - .40 mol - xs mol +.40 mol +.40 mol ____________________________________After 0mol xsmol .40 mol .40 mol • Excess is written (xs) and indicates there is some reactantremaining. Here, PbS is completelyconsumedand some O2remains after the reaction is complete.

5. Further Reality • What if only a certain amount of each reactant were available? • 25.50 g of oxygen reacts with 114.85 g lead (II) sulfide producing lead (II) oxide and sulfur dioxide. What mass of lead (II) oxide would be produced? • We cannot directly measuremoles, so the reactant amounts are given in grams. In order to use our BCA table (for moleratios) we need the amounts in moles. Using molar mass, we convert the mass of the reactants to moles of reactants.

6. Mass To Moles, “Molar Mass” • 25.50 g O2 x 1 mol O2_ = .80 mol O2 32.0 g O2 • 114.85 g PbS x _1 mol PbS_ = .48 mol PbS 239.27g PbS

7. A Second Look • Equation:2PbS + 3O2 2PbO + 2SO2Before: .48 mol .80 mol 0 mol 0 molChange -__mol - __mol +__mol+__mol _______________________________ • After __mol __mol __mol __ mol

8. Limiting & Excess Reactants • Our first task is to find out which reactant will be completely consumed(limitingreactant) and which reactant will have some remaining after the reaction is complete (reactant in excess). We will use moleratiosfrom the BCA table for this task.

9. Limiting & Excess Reactants • .80 mol O2 x 2 mol PbS = .53 mol PbS 3 mol O2 We need .53 mol of PbS to completely react .80 mol of O2 • .48 mol PbS x 3 mol O2 = .72 mol O2 2 mol PbS We need .72 mol of O2 to completely react .48 mol of PbS

10. Evaluating Our Answers • We need 0.53 mol PbS to completely burn 0.80 mol O2. We only have 0.48 mol PbS. Not all of the O2 will be “consumed”.The reaction will stop when the PbS has run out. This tells us the PbS will limit the reaction (limiting reactant) and some oxygen will remain after the reaction is complete (reactant in excess).

11. Amount of Reactant Remaining • We need 0.72 mol O2 to completely react with 0.48 mol PbS. We have 0.80 mol O2.0.08 mol O2 will remainafter all of the PbS has been consumed.

12. Importance of Limiting Reactant • The PbS limitsthe reaction. PbS “runs out” before all the O2 is consumed. PbS is the reactant that determineshow much product will be produced.

13. Putting It All Together! • Equation: 2PbS + 3O2 2PbO + 2SO2Before: .48 mol .80 mol 0 mol 0 mol Change-.48 mol - .72 mol + .48 mol + .48 mol ___________________________________After:0mol .08 mol .48 mol .48 mol