The document discusses reaction stoichiometry and determining limiting and excess reactants. It provides examples to illustrate: 1) When given amounts of multiple reactants, you cannot assume you can simply add the amounts, as one reactant may be limiting. 2) The limiting reactant is the one that is completely consumed in the reaction, while excess reactants remain afterward. 3) Worked examples show how to calculate the limiting reactant by converting masses to moles and identifying which is the smaller amount, and using the limiting reactant to calculate the mass of products formed.

1. Reaction StoichiometryDetermining the Limiting and Excess Reactants

2. Reaction Stoichiometry: Limiting and Excess Reactant IWhen given the amounts of two or more reactants, can not assume to simply add the amounts given. Example: If given 10 grams of sodium, and 10 grams of chlorine, would NOT produce 20.0 grams of sodium chloride.Na + Cl2  NaCl 10.0g 10.0g \ 20.00g

3. Reaction Stoichiometry: Limiting and Excess Reactant IIWe can not add the amounts because need to see which reactant LIMITS the amount of products to be produced. Limiting Reactant: the reactant which is completely used (or consumed) in the chemical reaction. Each reaction can only have ONE limiting reactant.The excess Reactant(s): The reactant or reactants that are NOT completely consumed in the chemical reaction. There may be MORE THAN ONE excess reactant.

4. Science, “Do Now” When given the following equation:2 wheels + 1 body + 1 handle bar + 1 gear chain  1 bikeDetermine how many bikes would you be able to produce with an inventory of: 50 gear chains 52 wheels 75 handle bars30 bodyHow many of each part is left over?

5. Science ‘Do Now’ Answer2 wheels + 1 body + 1 handle bar + 1 gear chain  1 bikeDetermine how many bikes would you be able to produce with an inventory of: 30 BIKES, limited by the number of bike body’s availableHow many of each part is left over?

6. Determining the Limiting and Excess Reactant – Bike ExampleWhich reactant was completely consumed (used?) Bike Body is the Limiting Reactant because there was nothing left over. Which reactant(s) were in excess? Gear Chains, Wheels AND Handle Bars because these reactants did have parts left over.

7. Example 1a: Identify the Limiting and Excess Reactant2H2 + O2 2H2O8.0g 8.0gStep 1: Identify the Limiting Reactant – go to moles Larger number = excess reactantSmaller number = limiting reactant=3.97 mol H2O8.0g H2 x 1 mol H2 2.02g H2x 2 mol H2O 2 mol H2=0.5 mol H2O8.0g O2 x 1 mol O2 32.00g O2x 2 mol H2O 1 mol O21 mol O2 = 32.0og O21 mol O2 = 2 mol H2O1 mol H2 = 2.02g H22 mol H2 = 1 mol H2OConversionKeys

8. Example 1b: Determine how much product will be produced in grams.2H2 + O2 2H2O8.0g 8.0g ?g Step 2: Use the limiting reactant mol to obtain product. x 18.02g H2O 1 mol H2O= 9.01g H2O0.5 mol H2O2 mol H2O = 1 mol O218.02g H2O= 1 mol H2OConversionKeys

9. 11-3 Worksheet #15 Part A2Al+ Fe2O3 2Fe + Al2O321.4g 91.3g Step 1: Identify the Limiting Reactant – go to moles 21.4g Al x 1 mol Al 26.98g Alx 2 mol Fe 2 mol AlExcessreactantLimitingreactant=0. 793 mol Fe91.3g Fe2O3 x 1 mol Fe2O3 159.7g Fe2O3x 2 mol Fe 1 mol Fe2O3=0.572 mol Fe1 mol Fe2O3 = 159.7g Fe2O31 mol Fe2O3 = 2 mol Fe1 mol Al = 26.98g Al2 mol Fe= 2 mol AlConversionKeys

10. 11-3 Worksheet #15 Part B2Al+ Fe2O3 2Fe + Al2O321.4g 91.3g ?gStep 2: Identify the Limiting Reactant – go to moles 0.572 mol Fex 55.84g Fe1 mol Fe= 31.9g Fe produced1 mol Fe= 55.84g FeConversionKeys

11. 11-3 Worksheet #7 Part ACaC2 + 2H2O 2Ca(OH) 2 + C2H243.25g 33.71g ?gStep 1: Identify the Limiting Reactant – go to moles LimitingreactantExcessreactant43.25g CaC2 x 1 mol CaC2 26.98g CaC2x 1 mol C2H2 1 mol CaC2=0. 674 mol C2H233.71g H2O x 1 mol H2O 18.02g H2Ox 1 mol C2H2 2 mol H2O=0.9353 mol C2H21 mol H2O = 18.02g H2O 2 mol H2O = 1 mol C2H21 mol CaC2 = 64.10g CaC21 mol CaC2 = 1 mol C2H2ConversionKeys

12. 11-3 Worksheet #7 Part BCaC2 + 2H2O 2Ca(OH) 2 + C2H243.25g 33.71g ?gStep 2: Use the limiting reactant mol to obtain product. 0. 674 mol C2H2x 26.04g C2H21 mol C2H2= 17.55g C2H2 produced1 mol C2H2= 26.04g C2H2ConversionKeys

13. 11-3 Worksheet #6 H2SO4 + 2NaOH Na2SO4 + 2H2O6.33g 5.92g ?g6.33g H2SO4 x 1 mol H2SO4 98.09g H2SO4x 1 molNa2SO4 1 mol H2SO4=0.064 mol Na2SO45.29g NaOH x 1 mol NaOH 40.00g NaOHx 1 mol Na2SO4 2 mol NaOH=0.074 mol Na2SO40.064 mol Na2SO4x 142.04g Na2SO4 1 mol Na2SO4= 9.09g Na2SO4 produced