1. Limiting Reagents and Percent Yield Using Stoichiometry

2. Chemistry Joke Q: Why were copper and titanium the cheerleader’s favorite elements? A: Because she knew she was a Cu Ti!

3. Limiting Reagent—t Limiting Reagent—the reactant that controls the quantity of product formed A real world example… (We use the terms reagent and reactant interchangeably.)

4. You want to make a lot of chocolate chip cookies! You have – 4 lbs of butter, 2 lbs of salt, 1 gallon of vanilla extract, 8 lbs of chocolate chips, 20 lbs flour, 15 lbs of sugar, 10 lbs of baking soda, and TWO eggs. –What is the limiting reagent???

5. Limiting Reactant – the reactant that is used up completely. Excess Reactant – the reactant that is not used up completely in a reaction. There’s some left over.

6. Finding the Limiting Reagent All limiting reagent problems will have more than one reactant amount. MUST If there is more than one given amount of reactant, you MUST find the limiting reagent.

7. Step 2 Divide each number of moles by the coefficient for that reactant in the balanced equation. Step 1 – –If the given amount of each reactant is not in moles-then convert each value into moles

8. Step 3 The smaller amount of reactant after dividing is the limiting reactant.

9. An Example… Zn + 2HCl  ZnCl 2 + H 2 What is the limiting reactant when 12.1 g Zn reacts with 2.65 g HCl? Notice the two “given” reactants.

10. 12.1 g Zn 65.38 g Zn 1 mol Zn= 0.185 mol Zn 1 mol Zn 2.65 g HCl 36.46 g HCl 1 mol HCl = 0.0727 mol HCl 2 mol HCl 0.0363 < 0.185 therefore, HCl is limiting. Change into moles. Divide by coefficient. Zn + 2HCl  ZnCl 2 + H 2 = 0.185 = 0.0363

11. Suppose 6.70 g Na reacts with 3.20 g Cl 2. Which is the limiting reactant? Suppose 6.70 g Na reacts with 3.20 g Cl 2. Which is the limiting reactant? 2 Na + Cl 2  2 NaCl Another Example… Change into moles. Divide by coefficient.

12. Finding the Amount of Product First, find the limiting reagent. MASS Then, use stoichiometry starting with the MASS of the LIMITING REAGENT!

13. For Example… A 50.6 g sample of Mg(OH) 2 is reacted with 45.0 g of HCl according to the reaction: Mg(OH) 2 + 2 HCl → MgCl 2 + 2 H 2 O A 50.6 g sample of Mg(OH) 2 is reacted with 45.0 g of HCl according to the reaction: Mg(OH) 2 + 2 HCl → MgCl 2 + 2 H 2 O How much MgCl 2 will be produced? There are 2 given reactants so I must first find the limiting one. I will then use the mass of the limiting reactant to find how much product will be produced.

14. For Example… A 50.6 g sample of Mg(OH) 2 is reacted with 45.0 g of HCl according to the reaction: Mg(OH) 2 + 2 HCl → MgCl 2 + 2 H 2 O A 50.6 g sample of Mg(OH) 2 is reacted with 45.0 g of HCl according to the reaction: Mg(OH) 2 + 2 HCl → MgCl 2 + 2 H 2 O How much MgCl 2 will be produced?

15. Percent Yield Theoretical Yield – the maximum, calculated amount of product that can be produced – –This number comes from stoichiometry!! Actual Yield – the measured amount of product (experimental yield) – –This number comes from lab experiments. Percent Yield = actual yield theoretical yield x 100

16. An Example… When 84.8 g of iron (III) oxide reacts with an excess of carbon monoxide, 54.3 g of iron is produced. What is the % yield? Fe 2 O 3 + 3 CO → 2 Fe + 3CO 2 Label the problem: Notice that one number in the problem is the amount of a reactant, and the other number in the problem is the actual yield of a product. 84.8 g 54.3 g

17. Use stoichiometry to find the theoretical yield in grams. Fe 2 O 3 + 3 CO → 2 Fe + 3CO 2 Start with the amount of reactant. The compound that was actually produced is your unknown. 84.8 g 54.3 g actual yield Then, divide the actual yield by your answer and multiply by 100 to find percent yield.

18. Chemistry Joke It’s weird that uranium and iodide don’t combine more in nature… Because I think U and I would be great together!!