DO NOW!!! Back of Worksheet! If 34.5 g of Copper reacts with 70.2 g of silver nitrate, according to the following reaction, what is the maximum number of grams of silver that can be produced? __ Cu + ___ Ag(NO3)  __Cu(NO3)2 + __ Ag

After today you will be able to… Identify the Limiting Reagent in a chemical equation Solve percent yield problems

Limiting Reactant Limiting Reactant When 2 substances are allowed to react one of them usually reacts completely while the other is not all used up because it is present in excess The reactant that is completely used up in the reaction is the limiting reactant. It determines the amount of products formed. The other reactant that is not completely consumed in the reaction is the excess reactant

Yield- amount of Product Actual yield -> amount of product formed from the actual chemical reaction and is usually less than the theoretical yield (this is given in the question) Theoretical yield  the maximum amount of products which could be produced by the complete reaction of the limiting reactant (based on stoichiometric calculations)

Yield: the amount of product How to Calculate! In chemical calculations, it is always assumed that the reaction goes to completion. However in reality, this seldom happens. The expected amount of product is not usually obtained. percent yield = actual yield x 100% theoretical yield

Sample Problem Sample problem 1 mol H2 2.02 g H2 x 2 mol H2O 2 mol H2 What is the % yield of H2O if 138 g H2O is produced from 16 g H2 and excess O2? Step 1: write the balanced chemical equation 2H2 + O2  2H2O Step 2: determine actual and theoretical yield. Actual is given, theoretical is calculated: 1 mol H2 2.02 g H2 x 2 mol H2O 2 mol H2 x 18.02 g H2O 1 mol H2O x # g H2O= 16 g H2 143 g = Step 3: Calculate % yield actual theoretical 138 g H2O 143 g H2O = % yield = x 100% x 100% 96.7% =

Sample Problem Sample problem 2 2 mol NH3 3 mol H2 x 17.04 g NH3 What is the % yield of NH3 if 40.5 g NH3 is produced from 20.0 mol H2 and excess N2? Step 1: write the balanced chemical equation N2 + 3H2  2NH3 Step 2: determine actual and theoretical yield. Actual is given, theoretical is calculated: 2 mol NH3 3 mol H2 x 17.04 g NH3 1 mol NH3 x # g NH3= 20.0 mol H2 227 g = Step 3: Calculate % yield actual theoretical 40.5 g NH3 227 g NH3 = % yield = x 100% x 100% 17.8% =

Put it all together now!! Put it all Together Now!! 2H2 + O2  2H2O What is the % yield of H2O if 58 g H2O are produced by combining 60 g O2 and 7.0 g H2? Hint: determine limiting reagent first 1 mol O2 32 g O2 x 2 mol H2O 1 mol O2 x 18.02 g H2O 1 mol H2O x # g H2O= 60 g O2 68 g = 1 mol H2 2.02 g H2 x 2 mol H2O 2 mol H2 x 18.02 g H2O 1 mol H2O x # g H2O= 7.0 g H2 62.4 g = actual theoretical 58 g H2O 62.4 g H2O = % yield = x 100% x 100% 92.9% =

Last Example! Last Example!! 70 grams of manganese dioxide is mixed with 3.5 moles of hydrochloric acid. How many grams of Cl2 will be produced from this reaction if the % yield for the process is 42%? MnO2 + 4HCI  MnCl2 + 2H2O + Cl2 Last Example!

# g Cl2= 1 mol MnO2 86.94 g MnO2 x 1 mol Cl2 1 mol MnO2 x 70.9 g Cl2 70 grams of manganese dioxide is mixed with 3.5 moles of hydrochloric acid. How many grams of Cl2 will be produced from this reaction if the % yield for the process is 42%? MnO2 + 4HCI  MnCl2 + 2H2O + Cl2 # g Cl2= 1 mol MnO2 86.94 g MnO2 x 1 mol Cl2 1 mol MnO2 x 70.9 g Cl2 1 mol Cl2 x 70 g MnO2 57.08 g Cl2 = 1 mol Cl2 4 mol HCl x 71 g Cl2 1 mol Cl2 x 3.5 mol HCl 62.13 g Cl2 = actual theoretical x g Cl2 57.08 g Cl2 = % yield = x 100% x 100% 42% = 42% x 57.08 g Cl2 100% = x g Cl2 24 g Cl2 =

Questions? Complete Worksheet! Reaction on

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