1. 9.5 – NOTES Limiting Reactants and Percent Yield

2. III. Limiting Reactants
A. Why do reactions stop? Are all reactions 100% efficient?
Reactions might stop for a number of reasons including: consumption of one or multiple reactants; occurrence of side reactions;
B. Calculating the product when a reactant is limited
Limiting reactant: limits the extent to which a reaction can proceed, thus determining the amount of product formed; the portion of all the other reactants that remain after the reaction stops = excess reactants;
Theoretical yield: the maximum amount of product that can be formed from the quantities of reactants given

3. To determine the limiting reactant:
1. Read the question and write the balanced equation.
2. Label all given amounts (usually 2 reactant masses).
3. Do stoichiometry twice, determining the amount of product (or of one product if there is more than one), starting with each given reactant.
4. Choose the smaller answer – that is the theoretical yield. The reactant that gives you this answer is the limiting reactant. The other reactant is the excess reagent.

4. Examples: a. If 5. 0 grams of nickel and 5
Examples: a. If 5.0 grams of nickel and 5.0 grams of chlorine gas are reacted completely, how many grams of product could be formed, theoretically?
Ni + Cl2  NiCl2 Given: 5.0 g Ni ( = 58.7g/mol) Find: g NiCl2 ( = g/mol)
5.0 g Cl2 ( =71.0g/mol)
5.0g Ni x 1 mol Ni x 1 mol NiCl2 x g NiCl2
58.7g Ni mol Ni 1 mol NiCl2 = 11g NiCl2
5.0g Cl2 x 1 mol Cl2 x 1 mol NiCl2 x g NiCl2
71.0 g Cl2 1 mol Cl mol NiCl2 = 9.1g NiCl2
It is possible to determine the limiting reactant from the mole ratio; however, most problems ask for the mass.

5. b. If you react 13. 5 grams of HCl with 14
b. If you react 13.5 grams of HCl with 14.8 grams of FeS, how many grams of H2S could be produced? 2HCl + FeS  H2S + FeCl2 Given: 13.5g HCl (= 36.5g/mol) Find: g H2S (= 33.1g/mol) 14.8g FeS (=87.9g/mol) 13.5g HCl x 1 mol HCl x 1 mol H2S x 33.1g H2S 36.5g HCl 2 mol HCl 1 mol H2S = 6.12g H2S 14.8g FeS x 1 mol FeS x 1 mol H2S x 33.1g H2S 87.9g FeS 1 mol FeS 1 mol H2S = 5.57g H2S

6. C. Why use an excess of a reactant?
Helps force the reaction to completion and can sometimes speed up a reaction
often the least expensive reactant is used as the excess reactant
think of burner – use excess air to create a blue flame which completely burns the fuel rather than fuel as the excess which creates a residue (carbon) on glassware.

7. To determine the quantity of excess:
1. Read the question and write the balanced equation.
2. Use stoichiometry, determine the quantity of reactant needed to completely consume the limiting reactant.
3. Subtract the quantity used from the quantity given.

8. Example: Determine the amount of HCl in excess using the above example
Example: Determine the amount of HCl in excess using the above example. 2HCl + FeS  FeCl2 + H2S 14.8g FeS x 1 mol FeS x 2 mol HCl x 36.5g HCl = 12.3g HCl (used) 87.9g FeS 1 mol FeS 1 mol HCl 13.5g – 12.3 = 1.2g excess

9. IV. Percent yield
How efficient was your process?
Percent yield = experimental yield x 100
theoretical yield
Most reactions never succeed in producing the predicted amount of product
Why experimental yields < theoretical yields?
not every reaction goes cleanly or completely
many reactions stop before all the reactants are used up, so the actual amount of product is less than expected
liquid reactants or products may adhere to the surface of containers or evaporate
solid product is always left behind on filter paper or lost in the purification process
products other than the intended ones may be formed by competing reactions, thus reducing the yield of the desired product.

10. Examples: 1. You reacted 2.78 grams of magnesium with excess sulfuric acid. You massed the resulting magnesium sulfate and found it had a mass of grams. What was your percentage yield?
Mg + H2SO4  MgSO4 + H2 Given: 2.78g Mg (=24.3g/mol) Find: g MgSO4 (=120.4g/mol)
3.45g MgSO4 produced
2.78g Mg x 1 mol Mg x 1 mol MgSO4 x g MgSO4
24.3g Mg mol Mg 1 mol MgSO4 =13.8g MgSO4
% yield =3.45g MgSO4
13.8 g x 100 = 25.0%

11. 2. If you react 7.82 grams of copper with excess silver nitrate solution, how many grams of silver would be produced if the reaction were 85% efficient?
Cu + 2AgNO3  Cu(NO3)2 + 2Ag Given: 7.82g Cu (63.5g/mol) Find: g Ag ( 108g/mol)
7.82g Cu x 1 mol Cu x 2 mol Ag x g Ag
63.5g Cu mol Cu mol Ag = 26.6g Ag
26.6g Ag x .85 = 22.6g Ag