Limiting Reactants and % Yield

Limiting Factors zLimiting factors are those factors which control the speed or the outcome of a reaction or a situation yHave you ever been behind a school bus in the morning? yYou can only go as fast as the bus in front of you. yThe bus one limiting factor preventing you from getting to school on time.

Limiting Reactants Ingredients for making “S’mores”: 2 graham crackers 1 marshmallow 1 Hershey bar How many “S’mores” can you make with 14 graham crackers, 5 marshmallows, and 6 Hershey bars? 5 “S’mores” How many of each ingredient are left over? 0 marshmallows 4 graham crackers 1 Hershey bar

Limiting Reactant zThe limiting reactant is the reactant which controls the rate of reaction or the amount of product that can be made zThe limiting reactant is determined using stoichiometric relationships

Excess zThe reactant that you have more than you need to perform a chemical reaction is called the excess reactant zIt is not completely used up in a chemical reaction.

Here’s an example zYou are making ham and cheese sandwiches and you have: y5 pieces of ham y5 pieces of cheese y8 pieces of bread zWhich of these ingredients do you have more than enough of (excess reactant)? zWhich of these do you not have enough of (limiting reactant?

Yield zYield is how much product that you produce zIn industry, the amount of yield needs to be calculated so that production schedules can be made zYou also need to know how much product you are going to be making so that you will have the appropriate size collection container

Real or Not? zActual yield refers to the amount of product that is actually generated zTheoretical yield refers to the amount of product that you expect to generate

Calculating Yield zIn order to calculate the theoretical yield you need to yWrite a balanced chemical equation. yCalculate the amount of product produced from each reactant individually. yThe reactant that produces the least amount of product is the limiting reactant. yThe reactant that you have enough of is the excess reactant.

Percentage Yield You recently bought a new car. The EPA sticker says that you should get 28 mpg in “average driving”. Your actual miles per gallon turns out to be less than 28 mpg. Why?

Theoretical Yield vs. Actual Yield You perform a mass- mass calculation to determine how much chemical should be produced in a reaction. The actual results of the experiment produce less chemical than calculated. Why?

Calculating Percent Yield 1.Calculate the theoretical yield (mass-mass calculation). 2.Determine the actual yield. 3.Calculate the % yield % Yield = Actual yield x 100 Theoretical yield

A Percent Yield Problem A chemist starts with 1.75 g of salicylic acid (C 7 H 6 O 3 ) and excess methanol (CH 3 OH) and reports the production of 1.42 g oil of wintergreen (C 8 H 8 O 3 ) in the following reaction: C 7 H 6 O 3 + CH 3 OH  C 8 H 8 O 3 + H 2 O What is the percent yield for this reaction?

Solving the Problem C 7 H 6 O 3 + CH 3 OH  C 8 H 8 O 3 + H 2 O g?g g x 1 mol = 138 g mol C 7 H 6 O mol C 7 H 6 O 3 x 1 molC 8 H 8 O 3 = 1 mol C 7 H 6 O mol C 8 H 8 O mol C 8 H 8 O 3 x 152 g = 1 mol 1.93 g C 8 H 8 O 3

% Yield cont. 6. (Actual Yield ÷ Theoretical Yield) x 100 (1.42 g ÷ 1.93 g) x 100 = 73.6 % Yield

Practice Problems 1.A chemist carried out a reaction that should produce 21.8 g of a product, according to a mass-mass calculation. However, the chemist was able to recover only 13.9 g of the product. What percentage yield did the chemist get? 2.A calculation indicates that 82.2 g of a product should be obtained from a certain reaction. If a chemist actually gets 30.7 g, what is the percentage yield? 3.Chromium(III) hydroxide will dissolve in sodium hydroxide according to the following equation: NaOH + Cr(OH) 3  NaCr(OH) 4 If you begin with 66.0 g of Cr(OH) 3 and obtain 38.4 g of product, what is your % yield?

Solutions 63.8% #1.(13.9g / 21.8g) x 100 = 63.8% 37.3% #2.(30.7g / 82.2g) x 100 = 37.3% #3.NaOH + Cr(OH) 3  NaCr(OH) g ?g 66.0g Cr(OH) 3 x 1mol/103g = 0.641mol Cr(OH) 3 = 0.641mol NaCr(OH) 4 x 143g / 1mol NaCr(OH) 4 = 91.66g NaCr(OH) % (38.4g / 91.7g) x 100 = 41.9%

Practice Problem Identify the limiting reactant and the theoretical yield of H 3 PO 3 if 225 g of PCl 3 is mixed with 125 g of H 2 O PCl 3 + 3H 2 O  H 3 PO 3 + 3HCl Convert each mass to moles: 225 g PCl 3 x 1 mol/137 g = 1.64 mol PCl g H 2 O x 1 mol/ 18 g = 6.94 mol H 2 O 1.64 mol PCl 3 requires 4.92 mol H 2 O PCl 3 is the limiting reactant 1.64 mol PCl 3 = 1.64 mol H 3 PO mol H 3 PO 3 x 82 g/1 mol = 134 g H 3 PO 3 225g 125g ?g

Classwork zP. 314 #1-3 zp. 317 #1-3 zp. 318 #1-3

Homework zP