Entry Task: March 13-14 th Block 2. Agenda: Discuss Limited and % yield ws Self Check on Limited reactant & % yields

Entry Task: March 13-14th Block 2

Agenda:

• Discuss Limited and % yield ws• Self Check on Limited reactant & % yields

I can…

• Use the sequence of steps used in solving stoichiometric problems

• Identify the limiting reactant in a chemical equation• Identify the excess reactant and calculate the amount

remaining after the reaction is complete.• Calculate the mass of a product when the amounts of

more that one reactant are given• Calculate the theoretical yield of a chemical reaction

from data.• Determine the percent yield for a chemical reaction.

Limited reactants % yields practice

A reaction of 25 grams of sodium bromide with 20.0 grams of potassium chloride creates sodium chloride and potassium bromide. How much sodium chloride would be produced

(theoretically)? Identify the limited reactant, the amount of excess in this reaction.

___NaBr + ___KCl ___KBr + ___NaCl


A reaction of 25 grams of sodium bromide with 20.0 grams of potassium chloride creates sodium chloride and potassium bromide. How much sodium chloride would be produced (theoretically)? Identify the limited reactant, the amount of excess in this reaction.

25.0 g NaBr 14 g NaCl

103 g NaBr

1 mole NaBr 58.4 g NaCl

1 mol NaCl1 mole NaBr

1 mol NaCl

20 g KCl 15.6 g NaCl

74.5g KCl

1 mole KCl 58.4 g NaCl

1 mole NaCl1 mole KCl

1 mole NaCl

Which reactant is Limited and which reactant is Excess?

Limited= NaBr

Excess= KCl


25.0 g NaBr 14 g NaCl

103 g NaBr



1 mol NaCl

20 g KCl 15.6 g NaCl

74.5g KCl



1 mole NaCl

A reaction of 25 grams of sodium bromide with 20.0 grams of potassium chloride creates sodium chloride and potassium bromide. How much sodium chloride would be produced (theoretically)? Identify the limited reactant, the amount of excess in this reaction.

Start with the limited reactant!!

UsedIn reaction

20 grams KCl minus 18 g KCl =

Given UsedIn reaction

2 gEXCESS KCl


25.0 g NaBr 18 g KCl

103 g NaBr

1 mole NaBr 74.5g KCl

1 mole KCl1 mole NaBr

1 mole KCl

Percent Yield= Actual yield (from experiment)

Theoretical yield (from stoich calculation) X 100

If 12 grams of sodium chloride were produced after the reaction, what is the % yield?

Percent Yield= 12 g (from experiment)

14 g (from stoich calculation-limit reactant) X 100

86%


With the same reaction, what if 50.0 grams of sodium bromide reacted with 75.0 grams of potassium chloride. How much sodium chloride would be produced (theoretically)? Identify the limited reactant, the amount of excess in this reaction.

50.0 g NaBr 28.3 g NaCl

103 g NaBr



1 mol NaCl

25.0 g KCl 19.6 g NaCl

74.5g KCl



1 mole NaCl

Which reactant is Limited and which reactant is Excess?

Limited= KCl

Excess= NaBr


50.0 g NaBr 28.3 g NaCl

103 g NaBr



1 mol NaCl

25.0 g KCl 19.6 g NaCl

74.5g KCl



1 mole NaCl

With the same reaction, what if 50.0 grams of sodium bromide reacted with 25.0 grams of potassium chloride. How much sodium chloride would be produced (theoretically)? Identify the limited reactant, the amount of excess in this reaction.

Start with the limited reactant!!

UsedIn reaction

50.0 grams NaBr minus 34.6 g NaBr =

Given UsedIn reaction

15.4 gEXCESS NaBr


25.0 g KCl 34.6 g NaBr

74.5g KCl

1 mole KCl 103 g NaBr

1 mole NaBr1 mole KCl

1 mole NaBr

Percent Yield= Actual yield (from experiment)

Theoretical yield (from stoich calculation) X 100

If 14 grams of sodium chloride were produced after the reaction, what is the % yield?

Percent Yield= 14 g (from experiment)

19.6 g (from stoich calculation-limit reactant) X 100

71%

25 grams of lead II iodide reacts with potassium nitrate to create lead II nitrate and potassium iodide.

___PbI2 + ___KNO3 ___Pb(NO3)2 + ___KI22

___PbI2 + _2_KNO3 ___Pb(NO3)2 + _2_KI

1. How much potassium nitrate is needed if 25 grams of lead II iodide were used in the reaction?

25.0 g PbI2 11.0 g KNO3

461 g PbI2

1 mole PbI2 101 g KNO3

1 mol KNO31 mole PbI2

2 mol KNO3

I needed to find out how much of the OTHER reactant I need to complete this reaction. Compare reactant to reactant!!

___PbI2 + _2_KNO3 ___Pb(NO3)2 + _2_KI

2. How much potassium iodide was produced from the reaction with 25 grams of lead II iodide used?

25.0 g PbI2 18.0 g KI461 g PbI2

1 mole PbI2 166 g KI

1 mol KI1 mole PbI2

2 mol KI

Now I’m finding out how much product- KI is produced with my 25 grams of PbI2.

Compare reactant to product!!

___PbI2 + _2_KNO3 ___Pb(NO3)2 + _2_KI

3. Using the amount of potassium nitrate, how much potassium iodide would be produced?

11.0 g KNO3 18.0 g KI101 g KNO3

1 mole KNO3 166 g KI

1 mol KI2 mole KNO3

2 mol KI

NOW I’m using the other reactant KNO3 amount that we figured out from the ratio with PbI2.

Compare reactant to reactant!!

___PbI2 + _2_KNO3 ___Pb(NO3)2 + _2_KI

4. What can you deduce from your answers from 2 and 3?

11.0 g KNO3 18.0 g KI101 g KNO3

1 mole KNO3 166 g KI

1 mol KI2 mole KNO3

2 mol KI

25.0 g PbI2 18.0 g KI461 g PbI2

1 mole PbI2 166 g KI

1 mol KI1 mole PbI2

2 mol KI

By reacting 25 g of PbI2 and 11 g of KNO3, you will produce 18 g of KI NO WASTE!!.

___PbI2 + _2_KNO3 ___Pb(NO3)2 + _2_KI

5. What if I need 100 grams of potassium nitrate to be produced? How much of each reactant would be needed to complete this reaction? Work backwards

100 g KNO3 164 g KI101 g KNO3

1 mol KNO3166 g KI

1 mol KI2 mol KNO3

2 mol KI

100 g KNO3 164 g Pb(NO3)2

101 g KNO3

1 mol KNO3331.12g Pb(NO3)2

1 mole Pb(NO3)22 mol KNO3

1 mole Pb(NO3)2

LETS CHECK OUR WORK!!!

___PbI2 + _2_KNO3 ___Pb(NO3)2 + _2_KI

5. What if I need 100 grams of potassium nitrate to be produced? How much of each reactant would be needed to complete this reaction? Work backwards

164 g KI 99.88 g KNO3

166 g KI

1 mol KI 101.1 g KNO3

1 mol KNO32 mol KI

2 mol KNO3

164 g Pb(NO3)2 100g KNO3

331.12g Pb(NO3)2

1 mol Pb(NO3)2

101 g KNO3

1 mole KNO31 mol Pb(NO3)2

2 mole KNO3

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Entry Task: March 13-14 th Block 2. Agenda: Discuss Limited and % yield ws Self Check on Limited reactant & % yields