Percentage Yield and Energy Lesson 3. Sometimes reactions do not go to completion. Reaction can have yields from 1% to 100%. 1.How many grams of Fe are.

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Presentation transcript:

Percentage Yield and Energy Lesson 3

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 Fe 2 O 3, if the percentage yield is 75.0%? = 52.4 g x 0.750x 55.8 g 1 mole x 4 mole Fe 2 mole Fe 2 O 3 x 1 mole g 100. g Fe 2 O g? g Fe 2 O C  Fe +3 CO 2 4 2

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 AgNO 3 is used to make 75.1 g of Ag 2 SO 4(s). Calculate the percentage yield. AgNO 3(aq) + Na 2 SO 4(aq)  Ag 2 SO 4(s) + 2NaNO 3(aq) 75.1 g actual yield ? g152 g = gx g 1 mole x 1 Ag 2 SO 4 2 mole AgNO 3 x 1 mole g 152. g AgNO 3 =53.8 %x 100 %= % yield 1 2

Energy Calculations The energy term in a balanced equation can be used to calculate the amount of energy consumed or produced in a reaction. 3.How much energy is required to produce 25.4 g of H 2 ? +2H 2 O  H 2 +O g? kJ = 1340 kJ 2 mole H 2 x 213 kJ 2.02 g x 1 mole25.4 g H kJ

4.How many molecules of H 2 can be produced when 452 kJ of energy if consumed? ? Molecules452 kJ = 2.55 x molecs x 6.02 x molecs 1 mole213 kJ x 2 moles H kJ kJ2 H 2 O + H2+O2H2+O2 

5.How much energy is produced by an explosion of a 5.2 L balloon full of hydrogen at STP? H 2 +O 2  2H 2 O + ? kJ5.2 L = 25 kJ x 213 kJ 2 moles H L x 1 mole5.2 L 213 kJ 2