1. Limiting Reactants Chapter 13

2. A real-life example Cookies anyone? Given the following recipe for chocolate chip cookies… What if we only have 1 egg? How many cookies could we make? What would we need to do? What would we have left over? 2 ¼ cups flour 1 tsp salt 1 tsp baking soda 1 c butter (2 sticks) ¾ cup brown sugar ¾ cup granulated sugar 2 eggs 2 tsp vanilla 3 oz chocolate chips Yield: 4 dozen cookies

3. Demo: Bolts & Washers 2Mg + O 2  2MgO 1 bolt = 1 mole of Mg 2 washers = 1 mole of O 2 What would 1 mole MgO look like? Let’s “make” our reaction with bolts and washers…

4. Demo: Bolts & Washers + = 2Mg + O 2 = 2MgO

5. Demo: Bolts & Washers When does that “reaction” stop? What did you run out of? –Liming Reactant What is left over? –Excess Reactant

6. Definitions Reactant –Is a substance that takes part in a chemical reaction to make a product Limiting Reactant –The reactant that is completely used up in the reaction and limits the amount of product Excess Reactant –The reactant that is in excess, not all used up.

7. Why do we care? (Don’t write…let’s discuss) Scientific Reasons –Determine how much reactant you need –Determine how much product will be produced Economic (Show me the money!) –Keep production cost’s minimal –Maximize output in production

8. Let’s Review… Steps in solving stoichiometry problems… (how do we find mass or volume of a product given mass or volume of a reactant?) 1.Write and Balance Equation 2.Identify known and unknown 3.Convert mass (or volume) to moles 4.Use mole ratio to convert from moles of known to moles of unknown 5.Convert moles of unknown to mass (or volume) of unknown

9. Steps in Solving LR problems! When two amounts of reactants are given in a problem, we need to identify the limiting reactant to solve for the amount of product possible! 1.Start with a Balanced Equation 2.Convert grams of each reactant to moles (this is how much you have of each) 3.Use mole ratio to convert moles of one react to moles of the other and compare to the original number of moles obtained in step two (what you need) 4.Decide which is the limiting reactant

10. Another Method 1.Always begin with a BALANCED EQUATION!! 2.Convert the amount (grams/volume) of each reactant to moles of reactants. 3.Convert the number of moles of reactant to moles of product (doesn’t matter which one as long as you use the same product for both reactants) 4.Which ever reactant produces the least product is the limiting reactant.

11. Example 1 The reaction begins with 2.51g of HF and 4.56g of SiO 2.The reaction begins with 2.51g of HF and 4.56g of SiO 2. HF + SiO 2  SiF 4 + H 2 O Balance EquationBalance Equation Convert g  mols of each (What you have)Convert g  mols of each (What you have) Pick one & convert mols  moles of other reactant. This is what you need…Do you have enough?Pick one & convert mols  moles of other reactant. This is what you need…Do you have enough? If “no”, then that is your Limiting Reactant… if yes than the other is your LRIf “no”, then that is your Limiting Reactant… if yes than the other is your LR

12. 2. Calculate moles of each reactant (what you have) 2.51 g HF 1 mol HF = 0.1255 mol HF 20.0 g HF 4.56 g SiO 2 1 mol SiO 2 = 0.059 mol SiO 2 76.05 g SiO 2 Example 1 1. Balance the Equation 4HF + SiO 2  SiF 4 + 2H 2 O

13. Example 1 3. Use mole ratio to convert from one reactant to the other. 0.1255 mol HF 1 mol SiO 2 = 0.031 mol SiO 2 4 mol HF Compare to moles of SiO 2 we have (from Step 2) 0.059 mol SiO 2 (have) > 0.031 mol SiO 2 (need) 4. Determine the limiting reactant If you have more than you need, this is not the limiting reactant This means the other reactant (HF) must be limiting

14. How much excess reactant is left over? –Determine the amount of the excess reactant used then subtract from the starting amount.

15. How much product can be formed? Start with the amount (mass or volume) of limiting reactant Use stoichiometry (mole ratio conversions) to find the amount (mass or volume) of product left over

16. Important Info Chapter 11 Homework:Chapter 11 Homework: –Red Book #23-27 Next Class:Next Class: –S’mores Lab! –Don’t forget to sign up and bring ingredients –Don’t forget to sign up and bring ingredients

17. Example 2 Determine the limiting reactant, if an 80.0g solution of NaOH, which is 40.0% NaOH by mass, reacts with a 75.0g solution of H 2 SO 4, which contains 45.0% water by mass. NaOH + H 2 SO 4  Na 2 SO 4 + H 2 O 1.Balance Equation 2.Calculate grams of NaOH and H 2 SO 4 3.g  mols of each (What you have) 4.Pick one & compare (What you need) 5.Do you have enough?

18. So…the reaction will go to completion until all the NaOH (LR) is used up. Calculate the amounts of each product that will be produced.

19. Example If 40.0 g of H 3 PO 4 react with 60.0 g of MgCO 3 calculate:If 40.0 g of H 3 PO 4 react with 60.0 g of MgCO 3 calculate: a.g of Mg 3 (PO 4 ) 2 produced b.g of CO 2 produced c.Volume of CO 2 at STP H 3 PO 4 + MgCO 3  Mg 3 (PO 4 ) 2 + CO 2 + H 2 O