Percentage Yield and Energy Lesson 3

Sometimes reactions do not go to completion Sometimes reactions do not go to completion. Reaction can have yields from 1% to 100%. 1. How many grams of Fe are produced by the reaction of 100. g of Fe2O3, if the percentage yield is 75.0%? 2 Fe2O3 + 3 C  Fe + 3 CO2 4 100. g ? g 100. g Fe2O3 x 1 mole x 4 mole Fe x 55.8 g x 0.750 = 52.4 g 159.6 g 2 mole Fe2O3 1 mole

Percentage Yield = Actual Yield x 100% Theoretical Yield   Actual Yield is what is experimentally measured. Theoretical Yield is what is calculated using stoichiometry.

2. In an experiment 152. g of AgNO3 is used to make 75. 1 g of 2. In an experiment 152. g of AgNO3 is used to make 75.1 g of Ag2SO4(s). Calculate the percentage yield.   AgNO3(aq) + Na2SO4(aq)  Ag2SO4(s) + 2NaNO3(aq) 75.1 g actual yield 2 1 152 g ? g 152. g AgNO3 x 1 mole x 1 Ag2SO4 x 311.9 g = 139.5 g 169.9 g 2 mole AgNO3 1 mole % yield = 75.1 x 100 % = 53.8 % 139.5

Energy Calculations The energy term in a balanced equation can be used to calculate the amount of energy consumed or produced in a reaction. How much energy is required to produce 25.4 g of H2? + 2H2O  H2 + O2 213 kJ 2 ? kJ 25.4 g 25.4 g H2 x 1 mole x 213 kJ = 1340 kJ 2.02 g 2 mole H2

4. How many molecules of H2 can be produced when 452 kJ of 4. How many molecules of H2 can be produced when 452 kJ of energy if consumed? 2 H2O + 213 kJ  2 H2 + O2 452 kJ ? Molecules 452 kJ x 2 moles H2 x 6.02 x 1023 molecs = 2.55 x 1024 molecs 213 kJ 1 mole

5. How much energy is produced by an explosion of a 5. 2 L 5. How much energy is produced by an explosion of a 5.2 L balloon full of hydrogen at STP?   H2 + O2  2H2O + 2 213 kJ 5.2 L ? kJ 5.2 L x 1 mole x 213 kJ = 25 kJ 22.4 L 2 moles H2