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III. Stoichiometry Stoy – kee – ahm –eh - tree. Chapter 12. Sections Click the section to jump to the slides. Mole Ratios Mole-to-Mole Calculations Mole-to-Mass Calculations Particle Calculations Molar Volume Calculations Limiting Reactants Percent Yields. Things you should remember.

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Iii stoichiometry stoy kee ahm eh tree

III. StoichiometryStoy – kee – ahm –eh - tree

Chapter 12


Sections click the section to jump to the slides
SectionsClick the section to jump to the slides

  • Mole Ratios

  • Mole-to-Mole Calculations

  • Mole-to-Mass Calculations

  • Particle Calculations

  • Molar Volume Calculations

  • Limiting Reactants

  • Percent Yields


Things you should remember
Things you should remember

  • From the Moles Unit:

    • Identify particles as atoms, molecules (mc), and formula units (fun)

    • 1 mole = 6.02 x 1023 atoms, molecules, or formula units

    • 1 mole substance = mass (in grams) from the periodic table

  • From the Naming & Formulas Unit:

    • How to write a formula given a chemical name

  • From the Chemical Reactions Unit:

    • How to write a chemical equation given words

    • Balancing equations


I stoichiometry
I. Stoichiometry

  • stoikheion, meaning element

  • metron, meaning measure

  • Thus Stoichiometry- measuring elements!


Iii stoichiometry
III. Stoichiometry

  • Stoichiometry (sometimes called reaction stoichiometry to specify its use in analyzing a chemical reaction) is the calculation of quantitative (numbers) relationships between the reactants and products in a balanced chemical reaction.


A basis for calculations
A. Basis for Calculations

  • The basis for properly working stoichiometry problems is the Balanced Chemical Equation and the Mole Ratio. These are VITAL TO YOUR SURVIVAL IN STOICHIOMETRY!!!


1 review of a balanced chemical equation
1. Review of a Balanced Chemical Equation

  • You must always check to ensure you have a proper chemical equation. It is not correctly balanced you can not use it for any calculations!!!!


Ex properly balance the following equation
Ex: Properly Balance the following equation

2

__H2SO4(aq) + __NaHCO3(s) 

__Na2SO4(aq) + __H2O(l) + __CO2(g)

2

2


2 molar ratio
2. Molar Ratio

  • Once the equation is properly balanced, there are relationships between all of the compounds involved. These are called MOLE RATIOS.

  • any two compounds can be written as a relationship in terms of moles


For the reaction h 2 so 4 2nahco 3 na 2 so 4 2h 2 o 2co 2
For the Reaction:H2SO4 + 2NaHCO3Na2SO4 + 2H2O + 2CO2

What’s the relationship between NaHCO3 and Na2SO4?

2 mole NaHCO3 = 1 mol Na2SO4

(it’s just the coefficients!!)

What’s the relationship between H2SO4 and CO2?

1 mol H2SO4 = 2 mol CO2

(it’s just the coefficients!!)


For the reaction h 2 so 4 2nahco 3 na 2 so 4 2h 2 o 2co 21
For the Reaction:H2SO4 + 2NaHCO3Na2SO4 + 2H2O + 2CO2

  • 1 mol H2SO4 = 2 mol NaHCO3

    or

  • 1 mc H2SO4 = 2 mc NaHCO3

    or

  • 1 mol/mc H2SO4

    2 mol/mc NaHCO3

    or

  • 2 mol/mc NaHCO3

    1 mole/mc H2SO4


For the reaction h 2 so 4 2nahco 3 na 2 so 4 2h 2 o 2co 22
For the Reaction:H2SO4 + 2NaHCO3Na2SO4 + 2H2O + 2CO2

  • 1 mol H2SO4 = 2 mol CO2

    or

  • 1 mc H2SO4 = 2 mc CO2

    or

  • 1 mol/mc H2SO4

    2 mol/mc CO2

    or

  • 2 mol/mc CO2

    1 mole/mc H2SO4


Example iron reacts with oxygen to create iron iii oxide
Example: Iron reacts with oxygen to create iron(III) oxide.

After writing a balanced equation, write down 3 possible relationships. (hint: just look at the coefficients.)

Fe + O2 Fe2O3

Skeletal equation:

4Fe + 3O2 2Fe2O3

Balanced equation:

4 mol Fe = 3 mol O2

Relationships:

4 mol Fe = 2 mol Fe2O3

3 mol O2 = 2 mol Fe2O3


Practice
Practice

1. Write three mole ratios (relationships) from the reaction below:

Al2S3 + H2O  Al(OH)3 + H2S

6

2

3

1 mol Al2S3 = 6 mol H2O

1 mol Al2S3 = 2 mol Al(OH)3

1 mol Al2S3 = 3 mol H2S

You should have 3

6 mol H2O = 2 mol Al(OH)3

6 mol H2O = 3 mol H2S

2 mol Al(OH)3 = 3 mol H2S


Practice1
Practice

1. Write three mole ratios (relationships) from the reaction below:

Al2S3 + H2O  Al(OH)3 + H2S

6

2

3

1 mol Al2S3 = 6 mol H2O

1 mol Al2S3 = 2 mol Al(OH)3

1 mol Al2S3 = 3 mol H2S

6 mol H2O = 2 mol Al(OH)3

6 mol H2O = 3 mol H2S


Practice2
Practice

2.Aluminum is produced by decomposing aluminum oxide into aluminum and oxygen.

a. Write a balanced equation.

b. Write all the molar ratios that can be derived from this equation.

2Al2O3  4Al + 3O2

2 mol Al2O3 = 4 mol Al

2 mol Al2O3 = 3 mol O2

4 mol Al = 3 mol O2


NOTE:

  • NOTE: Every time you do a stoichiometric calculation you MUST use a mole ratio. The mole ratio allows you to compare one compound in an equation with another. Don’t forget to balance your chemical equations!!!



B calculating problems stoich it up
B. Calculating Problems - Stoich it up!

  • Before any stoich problem you have to set it up. Consider this the pre-game warm-up. This should become second nature to you.


Pre game warm up
Pre-game Warm-up:

1. Write a balanced reaction.

2. Determine your given & want

3. Determine your relationships. If you see…

  • 2 different substances, determine their mole ratio

  • mass (g, mg, kg), calculate the molar mass of that substance

  • atoms, molecules, or fun, 1 mol = 6.02 x 1023of that type

  • extras like mg or kg, you know what to do

    Now you are ready to solve- IT’S GAME TIME.


Game time
Game Time:

  • put your GIVEN OVER 1

  • place your relationships where the units cancel out diagonally

  • everything equal to each other goes above and below each other

  • cancel out your units until you are left over with your wanted



For most of the examples we will be using this equation
For most of the examples we will be using this equation: our problems

N2 (g) + 3H2 (g)  2NH3 (g)

Haber Process: an industrial process for producing ammonia from nitrogen and hydrogen by combining them under high pressure in the present of an iron catalyst

source: worldnet.princeton.edu

Haber Process:


1 moles to moles

1. Moles to Moles our problems


Ex 1: How many moles of ammonia are produced from 4.00 moles of hydrogen gas in the presence of excess nitrogen?

Balanced Equation:

N2 (g) + 3H2 (g)  2NH3 (g)

No grams, No molar mass!!

Given :

Want :

Relationships:

4.00 mol H2

No fun, mc, No 6.02 x 1023!!

? mol NH3

2 mol NH3 = 3 mol H2

4.00 mol H2

2 mol NH3

= mol NH3

2.67

x

3 mol H2

1


Ex 2 how many moles of nitrogen were used if 7 8 moles of ammonia were made in excess hydrogen gas
Ex 2: How many moles of nitrogen were used if 7.8 moles of ammonia were made in excess hydrogen gas?

Balanced Equation:

N2 (g) + 3H2 (g)  2NH3 (g)

No grams, No molar mass!!

Given :

Want :

Relationships:

7.8 mol NH3

No fun, mc, No 6.02 x 1023!!

? mol N2

2 mol NH3 = 1 mol N2

7.8 mol NH3

1 mol N2

= mol N2

3.9

X

2 mol NH3

1


Ex 3 how many moles of hydrogen react with 13 moles of nitrogen
Ex 3: How many moles of hydrogen react with 13 moles of nitrogen?

Balanced Equation:

N2 (g) + 3H2 (g)  2NH3 (g)

No grams, No molar mass!!

Given :

Want :

Relationships:

13 mol N2

No fun, mc, No 6.02 x 1023!!

? mol H2

1 mol N2 = 3 mol H2

13 mol N2

3 mol H2

= moles H2

39

x

1 mol N2

1



Steps to success
Steps to Success! nitrogen?

1. Write a balanced reaction.

2. Determine your given & want

3. Determine your relationships. If you see…

  • 2 different substances, determine their mole ratio

  • mass (g, mg, kg), calculate the molar mass of that substance

  • atoms, molecules, or f.un, 1 mol = 6.02 x 1023of that type

  • extras like mg or kg, you know what to do


Ex 1 how many grams of ammonia are produced from 4 00 moles of hydrogen gas in excess nitrogen
Ex 1: How many grams of ammonia are produced from 4.00 moles of hydrogen gas in excess nitrogen?

Balanced Equation:

N2 (g) + 3H2 (g)  2NH3 (g)

How do I know when to use mole ratios, molar mass or 6.02 x 1023?

Given :

Want :

Relationships:

4.00 mol H2

? g NH3

2 mol NH3 = 3 mol H2

1 mol NH3 = 17.031g NH3

Look at the “given” and “want” for clues

No fun, mc, No 6.02 x 1023!!


Ex 1 how many grams of ammonia are produced from 4 moles of hydrogen gas in excess nitrogen
Ex 1: How many grams of ammonia are produced from 4 moles of hydrogen gas in excess nitrogen?

Balanced Equation:

N2 (g) + 3H2 (g)  2NH3 (g)

Given :

Want :

Relationships:

4.00 mol H2

? g NH3

2 mol NH3 = 3 mol H2

1 mol NH3 = 17.031g NH3

4.00 mol H2

2 mol NH3

17.031g NH3

x

x

3 mol H2

1 mol NH3

1

= g NH3

45.4


Ex 2 how many moles of ammonia are produced from 4 00 grams of hydrogen gas in excess nitrogen
Ex 2: How many moles of ammonia are produced from 4.00 grams of hydrogen gas in excess nitrogen?

Balanced Equation:

N2 (g) + 3H2 (g)  2NH3 (g)

Given :

Want :

Relationships:

4.00 g H2

? mol NH3

2 mol NH3 = 3 mol H2

1 mol H2 = 2.016g H2

Remember your Steps to Success!

No fun, mc, No 6.02 x 1023!!


Ex 2 how many moles of ammonia are produced from 4 grams of hydrogen gas in excess nitrogen
Ex 2: How many moles of ammonia are produced from 4 grams of hydrogen gas in excess nitrogen?

Balanced Equation:

N2 (g) + 3H2 (g)  2NH3 (g)

Given :

Want :

Relationships:

4.00 g H2

? mol NH3

2 mol NH3 = 3 mol H2

1 mol H2 = 2.016g H2

4.00 g H2

1 mol H2

2 mol NH3

x

x

2.016g H2

3 mol H2

1

= mol NH3

1.32


Practice3
Practice hydrogen gas in excess nitrogen?

1. How many grams of nitrogen will react with 3.40 moles of hydrogen to produce ammonia?

2. How many grams of hydrogen are required to make 54.0 moles of ammonia in excess nitrogen?

3. How many moles of nitrogen react completely with 3.70 moles of hydrogen?

31.7 g N2

163 g H2

1.23 mol N2


3 mass to mass

3. Mass to Mass hydrogen gas in excess nitrogen?


Ex 1 how many grams of ammonia are produced from 4 00 grams of hydrogen gas in excess nitrogen
Ex 1: How many grams of ammonia are produced from 4.00 grams of hydrogen gas in excess Nitrogen?

Balanced Equation:

N2 (g) + 3H2 (g)  2NH3 (g)

Given :

Want :

Relationships:

4.00 g H2

? g NH3

2 mol NH3 = 3 mol H2

1 mol H2 = 2.016g H2

Remember your Steps to Success!

1 mol NH3 = 17.031 g NH3


Ex 1 how many grams of ammonia are produced from 4 grams of hydrogen gas in excess nitrogen
Ex 1: How many grams of ammonia are produced from 4 grams of hydrogen gas in excess Nitrogen?

Balanced Equation:

N2 (g) + 3H2 (g)  2NH3 (g)

Given :

Want :

Relationships:

4.00 g H2

? g NH3

2 mol NH3 = 3 mol H2

1 mol H2 = 2.016g H2

1 mol NH3 = 17.031g NH3

17.031g NH3

1 mol H2

2 mol NH3

4.00 g H2

x

x

x

2.016g H2

3 mol H2

1 mol NH3

1

= g NH3

22.5


Ex 2 how many grams of nitrogen react with 12 0 grams of hydrogen
Ex 2: How many grams of Nitrogen react with 12.0 grams of Hydrogen?

Balanced Equation:

N2 (g) + 3H2 (g)  2NH3 (g)

Given :

Want :

Relationships:

12.0 g H2

? g N2

1 mol H2 = 2.016g H2

Remember your Steps to Success!

1 mol N2 = 28.014 g N2


Ex 2 how many grams of nitrogen react with 12 0 grams of hydrogen1
Ex 2: How many grams of Nitrogen react with 12.0 grams of Hydrogen?

Balanced Equation:

N2 (g) + 3H2 (g)  2NH3 (g)

Given :

Want :

Relationships:

12.0 g H2

? g N2

1 mol N2 = 3 mol H2

1 mol H2 = 2.016g H2

1 mol N2 = 28.014 g N2

28.014g N2

12.0 g H2

1 mol H2

1 mol N2

x

x

x

2.016g H2

1

3 mol H2

1 mol N2

= g N2

55.6


Practice4
Practice Hydrogen?

  • Determine the mass of NH3 produced from 280 g of N2.

  • What mass of nitrogen is needed to produce 100. kg of ammonia?

340 g NH3

8.22 x 104 g N2



Ex 1 how many grams of ammonia are produced from 2 09 x 10 14 mc of hydrogen gas
Ex 1: How many grams of ammonia are produced from 2.09 x 10 Hydrogen?14 mc of hydrogen gas?

Balanced Equation:

N2 (g) + 3H2 (g)  2NH3 (g)

Given :

Want :

Relationships:

2.09 x 1014 mc H2

? g NH3

2 mol NH3 = 3 mol H2

1 mol NH3 = 17.031g NH3

1 mol H2 = 6.02 x 1023 mc H2


Ex 1: How many grams of ammonia are produced from 2.09 x 10 Hydrogen?14 mc of hydrogen gas?

Balanced Equation:

N2 (g) + 3H2 (g)  2NH3 (g)

Given :

Want :

Relationships:

2.09 x 1014 mc H2

? g NH3

2 mol NH3 = 3 mol H2

1 mol H2 = 6.02 x 1023 mc H2

1 mol NH3 = 17.031g NH3

1 mol H2

2 mol NH3

2.09 x 1014 mc H2

17.031g NH3

x

x

x

3 mol H2

1

6.02 x 1023 mc H2

1 mol NH3

3.94 x 10-9

= g NH3


Ex 2 how many molecules of ammonia are produced from 40 2 g of h 2 in the presence of excess n 2
Ex 2: How many molecules of ammonia are produced from 40.2 g of H2 in the presence of excess N2?

Balanced Equation:

N2 (g) + 3H2 (g)  2NH3 (g)

Given :

Want :

Relationships:

40.2g H2

? mc NH3

2 mol NH3 = 3 mol H2

1 mol H2 = 2.016g H2

1 mol NH3 = 6.02 x 1023 mc NH3


Ex 2 how many molecules of ammonia are produced from 40 2 g of h 2 in the presence of excess n 21
Ex 2: How many molecules of ammonia are produced from 40.2 g of H2 in the presence of excess N2?

Balanced Equation:

N2 (g) + 3H2 (g)  2NH3 (g)

Given :

Want :

Relationships:

40.2g H2

? mc NH3

2 mol NH3 = 3 mol H2

1 mol H2 = 2.016g H2

1 mol NH3 = 6.02 x 1023 mc NH3

1 mol H2

40.2g H2

2 mol NH3

6.02 x 1023 mc NH3

x

x

x

2.016g H2

3 mol H2

1

1 mol NH3

8.00 x 1024

= mc NH3


Practice5
Practice of H

  • How many mc of nitrogen will react with exactly 7.04 grams of hydrogen?

  • How many mc of ammonia will 2.09 x 1021 mc of N2 produce in excess hydrogen?

7.01 x 1023 mc N2

4.18 x 1021 mc NH3


1 how many mc of nitrogen will react with exactly 7 04 grams of hydrogen
1) How many mc of nitrogen will react with exactly 7.04 grams of hydrogen?

Balanced Equation:

N2 (g) + 3H2 (g)  2NH3 (g)

Given :

Want :

Relationships:

7.04 g H2

? mc N2

1 mol N2 = 3 mol H2

1 mol H2 = 2.016g H2

1 mol N2 = 6.02 x 1023 mc N2

1 mol H2

7.04 g H2

1 mol N2

6.02 x 1023 mc N2

x

x

x

2.016g H2

3 mol H2

1

1 mol N2

7.01 x 1023

= mc N2


How many mc of ammonia will grams of hydrogen?2.09 x 1021 mc of N2 produce in excess hydrogen?

Balanced Equation:

N2 (g) + 3H2 (g)  2NH3 (g)

Given :

Want :

Relationships:

2.09 x 1013 mc N2

? mc NH3

2 mol NH3 = 1 mol N2

1 mol N2 = 6.02 x 1023 mc N2

1 mol NH3 = 6.02 x 1023 NH3

2.09 x 1013 mc N2

1 mol N2

2 mol NH3

6.02 x 1023 mc NH3

x

x

x

1

1 mol N2

1 mol NH3

6.02 x 1023 mc N2

4.18 x 1021

= mc NH3



4 molar volume of gases liters1
4. Molar Volume of Gases (liters) grams of hydrogen?

  • There is one new conversion: 1 mol of an ideal gas = 22.4 L of an ideal gas. For now we will assume every gas is an ideal gas at STP. You follow the same steps we have been following all along to solve this problem. In a reaction involving gases, Avogadro found that their volumes combine in whole number ratios. In other words, the coefficients in the chemical reaction also represent volume.


For gases when you see liters
For gases: grams of hydrogen?When you see liters…

1 mol __ = 22.4 L __


Ex 1: How many moles of nitrogen gas are needed to react with 44.8 liters of hydrogen gas to produce ammonia gas?

Balanced Equation:

N2 (g) + 3H2 (g)  2NH3 (g)

Given :

Want :

Relationships:

44.8 L H2

? mol N2

1 mol N2 = 3 mol H2

1 mol H2 = 22.4 L H2

Remember your Steps to Success!

No grams, No molar mass

No fun, mc, No 6.02 x 1023!!


Ex 1: How many moles of nitrogen gas are needed to react with 44.8 liters of hydrogen gas to produce ammonia gas?

Balanced Equation:

N2 (g) + 3H2 (g)  2NH3 (g)

Given :

Want :

Relationships:

44.8 L H2

? mol N2

1 mol N2 = 3 mol H2

1 mol H2 = 22.4 L H2

44.8 L H2

1 mol H2

1 mol N2

= mol N2

0.667

x

x

22.4 L H2

3 mol H2

1

Given over 1

Mole ratio

1 mol = 22.4 L


Ex 2 if 5 00 moles of h 2 react with excess n 2 how many liters of nh 3 are produced
Ex 2: If 5.00 moles of H with 44.8 liters of hydrogen gas to produce ammonia gas?2 react with excess N2, how many liters of NH3 are produced?

N2 (g) + 3H2 (g)  2NH3 (g)

Given :

Want :

Relationships:

5.00 mol H2

? L NH3

2 mol NH3 = 3 mol H2

1 mol NH3 = 22.4 L NH3

Remember your Steps to Success!

No grams, No molar mass

No fun, mc, No 6.02 x 1023!!


Ex 2: If 5.00 moles of H with 44.8 liters of hydrogen gas to produce ammonia gas?2 react with excess N2, how many liters of NH3 are produced?

N2 (g) + 3H2 (g)  2NH3 (g)

Given :

Want :

Relationships:

5.00 mol H2

? L NH3

2 mol NH3 = 3 mol H2

1 mol NH3 = 22.4 L NH3

2 mol NH3

5.00 mol H2

22.4 L NH3

= L NH3

74.7

x

x

3 mol H2

1 mol NH3

1

Given over 1

1 mol = 22.4 L

Mole ratio


Ii limiting reactants and excess reactants

II. Limiting Reactants and with 44.8 liters of hydrogen gas to produce ammonia gas?Excess Reactants


  • In reality, a scientist does not always add chemicals in perfect proportions. There is usually one reactant that is in “excess”.

  • The limiting reactant or limiting reagent is the reactant that limits the amounts of the other reactants that can combine and the amount of product that can form in a chemical reaction.

“which reactant are you going to run out of first?”


“which reactant do you have extra of?”


4 is sometimes called the

+

4

13

Which is the limiting reactant

4 shells

1 car

= 4 cars can be made if 4 shells are available

1 shell

13 tires

1 car

= 3.25 cars can be made if 13 tires are available

4 tires

Less product can be produced, therefore tires must be the limiting reactant


Activity
Activity is sometimes called the

  • Write a recipe for the perfect burger


Ex. Consider the following balanced equation: Zn(s) + 2HCl(aq)  ZnCl2(aq) + H2(g)If 1.21 moles of zinc are added to 2.64 moles of HCl, which reactant is in excess, and which is the limiting reactant?

Available

Needed

1.21 mol Zn

2 mol HCl

= 2.42 mol HCl

1 mol Zn

2.64 mol HCl

1 mol Zn

= 1.32 mol Zn

2 mol HCl


Ex. Consider the following balanced equation: Zn(s) + 2HCl(aq)  ZnCl2(aq) + H2(g)If 1.21 moles of zinc are added to 2.64 moles of HCl, which reactant is in excess, and which is the limiting reactant?

Available VS Needed

<

1.21 mol Zn 1.32 mol Zn

Limiting reactant

>

2.64 mol HCl 2.42 mol HCl

Excess reactant


To find the limiting reactant and the excess reactant
To find the limiting reactant and the excess reactant: 2HCl(aq)

1) Determine the balanced chemical equation

2) Convert the given/available amounts reactants to the number of moles of the other reactant(s)

3) Compare the available amounts to the needed amounts:

*If the available amount > needed amount, it is excess reactant

* If the available amount < needed amount, it is the limiting reactant

4) To calculate the amount of excess:

-Available Amount – Needed Amount = Amount of Excess

5) To calculate the amount of product produced:

-G: amount of available limiting reactant

-W: amount of product


Example
Example 2HCl(aq)

Some rocket engines use a mixture of hydrazine, N2H4, and hydrogen peroxide, as the propellant. The products formed are nitrogen gas and steam. Which is the limiting reactant when 0.75 mol of N2H4 is mixed with 0.50 mol of H2O2? How much excess reactant, in moles, remains unchanged? How much of each product, in moles, is formed?


Step 1: Write a balanced reaction 2HCl(aq)

Some rocket engines use a mixture of hydrazine, N2H4, and hydrogen peroxide, as the propellant. The products formed are nitrogen gas and steam.

N2H4

+ H2O2

2

4

N2(g)

+ H2O(g)


N 2HCl(aq) 2H4

+ H2O2

2

4

N2(g)

+ H2O(g)

  • Which is the limiting reactant when 0.750 mol of N2H4 is mixed with 0.500 mol of H2O2?

Available

Needed

Limiting Reactant

0.75 mol N2H4

2 mol H2O2

= 1.50 mol H2O2

1 mol N2H4

Excess

0.50 mol H2O2

1 mol N2H4

= 0.25 mol N2H4

2 mol H2O2


To find the limiting reactant and the excess reactant1
To find the limiting reactant and the excess reactant: 2HCl(aq)

4. To calculate the amount of excess:

-Available Amount – Needed Amount = Amount of Excess


Available 2HCl(aq)

Needed

  • How much excess reactant, in moles, remains unchanged?

0.75 mol N2H4

2 mol H2O2 =

1.50 mol H2O2

1 mol N2H4

0.50 mol H2O2

1 mol N2H4 =

Excess

0.25 mol N2H4

2 mol H2O2

-

= 0.50 mol N2H4

*Available – Needed = Remaining excess


To find the limiting reactant and the excess reactant2
To find the limiting reactant and the excess reactant: 2HCl(aq)

5. To calculate the amount of product produced:

-G: amount of available limiting reactant

-W: amount of product


N 2HCl(aq) 2H4

+ H2O2

2

4

N2(g)

+ H2O(g)

G:W:R:

0.50 mol H2O2

How many moles of product are formed?

G: amount of available limiting reactant

mole N2

N2 is the first product

2 mol H2O2 = 1 mol N2

0.50 mol H2O2

1 mol N2

= 0.25 mol N2

2 mol H2O2

G:W:R:

0.50 mol H2O2

G: amount of available limiting reactant

H2O is the second product

mole H2O

2 mol H2O2 = 4 mol H2O

0.50 mol H2O2

4 mol H2O

= 1.0 mol H2O

2 mol H2O2


Ex. Zinc metal and solid sulfur (S 2HCl(aq) 8) react to form solid zinc sulfide. If 2.0 moles of zinc are heated with 2.0 mol S8, identify the limiting reactant. How many moles of excess reactant remain. How many moles of product are formed?

Balanced Equation: 8 Zn + S8 8 ZnS

Available

Needed

2.0 mol Zn

1 mol S8

= 0.25 mol S8

8 mol Zn

1.0 mol S8

8 mol Zn

= 8.0 mol Zn

1 mol S8


Ex. Zinc metal and solid sulfur (S 2HCl(aq) 8) react to form solid zinc sulfide. If 2.0 moles of zinc are heated with 1.0 mol S8, identify the limiting reactant. How many moles of excess reactant remain. How many moles of product are formed?

Available VS Needed

<

2.0 mol Zn 8.0 mol Zn

Limiting reactant

>

1.0 mol S8 0.25 mol S8

Excess reactant


Ex. Zinc metal and solid sulfur (S 2HCl(aq) 8) react to form solid zinc sulfide. If 2.0 moles of zinc are heated with 1.0 mol S8, identify the limiting reactant. How many moles of excess reactant remain. How many moles of product are formed?

How much excess is present?

(Available Amount – Needed Amount = Amount of Excess)

1.0 mol S8 – 0.25 mol S8 = 0.75 mol S8 in excess


To find the limiting reactant and the excess reactant3
To find the limiting reactant and the excess reactant: 2HCl(aq)

4. To calculate the amount of excess:

-Available Amount – Needed Amount = Amount of Excess

5. To calculate the amount of product produced:

-G: amount of available limiting reactant

-W: amount of product


Ex. Zinc metal and solid sulfur (S 2HCl(aq) 8) react to form solid zinc sulfide. If two moles of zinc are heated with 1.00 mol S8, identify the limiting reactant. How many moles of excess reactant remain. How many moles of product are formed?

Balanced Equation: 8 Zn + S8 8 ZnS

How much product was formed?

G:W:R:

2 mol Zn

mol ZnS

8 mol Zn = 8 mol ZnS

2 mol Zn

8 mol ZnS

= 2 mol ZnS formed

8 mol Zn


Practice 2HCl(aq)

1. Carbon reacts with steam at high temperatures to produce hydrogen and carbon monoxide. If 2.4 mol of carbon are exposed to 3.1 mol of steam, identify the limiting reactant. How many moles of each product are formed?

Balanced Equation: C + H2O  H2 + CO

Available

Needed

2.4 mol C

1 mol H2O

= 2.4 mol H2O

1 mol C

3.1 mol H2O

1 mol C

= 3.1 mol C

1 mol H2O


Practice 1. Carbon reacts with steam at high temperatures to produce hydrogen and carbon monoxide. If 2.4 mol of carbon are exposed to 3.1 mol of steam, identify the limiting reactant. How many moles of each product are formed?

Available VS Needed

<

2.4 mol C 3.1 mol C

Limiting reactant

>

3.1 mol H2O 2.4 mol H2O

Excess reactant


To find the limiting reactant and the excess reactant4
To find the limiting reactant and the excess reactant: produce hydrogen and carbon monoxide. If 2.4 mol of carbon are exposed to 3.1 mol of steam,

4. To calculate the amount of excess:

-Available Amount – Needed Amount = Amount of Excess

5. To calculate the amount of product produced:

-G: amount of available limiting reactant

-W: amount of product


Balanced Equation: C + H produce hydrogen and carbon monoxide. If 2.4 mol of carbon are exposed to 3.1 mol of steam, 2O  H2 + CO

G:W:R:

2.4 mol C

Carbon reacts with steam at high temperatures to produce hydrogen and carbon monoxide. If 2.4 mol of carbon are exposed to 3.1 mol of steam, identify the limiting reactant. How many moles of each product are formed?

G: amount of available limiting reactant

mole H2

H2 is the first product

1 mol C = 1 mol H2

2.4 mol C

1 mol H2

= 2.4 mol H2

1 mol C

G:W:R:

2.4 mol C

G: amount of available limiting reactant

CO is the second product

mole CO

1 mol C = 1 mol CO

2.4 mol C

1 mol CO

= 2.4 mol CO

1 mol C


Set vi reactions
Set VI: Reactions produce hydrogen and carbon monoxide. If 2.4 mol of carbon are exposed to 3.1 mol of steam,

2) Zn + Pb(NO3)2 Pb + Zn(NO3)2

3) Fe + 2HCl  H2 + FeCl2


Iii theoretical yield actual yield and percent yield

III. Theoretical yield, Actual yield and Percent yield produce hydrogen and carbon monoxide. If 2.4 mol of carbon are exposed to 3.1 mol of steam,


C theoretical yield actual yield and percent yield
C. Theoretical yield, Actual yield and Percent yield produce hydrogen and carbon monoxide. If 2.4 mol of carbon are exposed to 3.1 mol of steam,

  • The yield of a chemical reaction is the quantity of product one obtains from a given ratio of reactants.

  • The actual yield of a chemical reaction is the mass of the compound that you actually recover when you are done with the reaction.

  • The actual yield is also referred to as the experimental yield

  • The theoretical yield is the mass of compound you should obtain (theoretically) if everything goes perfectly. In all of the examples above we have been pretending that everything is perfect. All of our wanteds have been theoretical.


Ex 1: What is the theoretical yield of Na produce hydrogen and carbon monoxide. If 2.4 mol of carbon are exposed to 3.1 mol of steam, 2SO4 in grams if 35 moles of NaOH is reacted with sufficient H2SO4?

  • Yield means product

  • Theoretical yield means mathematically, what should you get? (this is why we do stoichiometry calculations)


Ex 1: What is the theoretical yield of Na produce hydrogen and carbon monoxide. If 2.4 mol of carbon are exposed to 3.1 mol of steam, 2SO4 in grams if 3.50 moles of NaOH is reacted with sufficient H2SO4?

2NaOH + H2SO4 Na2SO4 + 2H2O

G:

W:

R:

3.50 mol NaOH

2 different substances

? g Na2SO4

You see g Na2SO4

2 mol NaOH = 1 mol Na2SO4

1 mol Na2SO4 = 142.042g Na2SO4

3.50 mol NaOH

1 mol Na2SO4

142.042g Na2SO4

x

x

2 mol NaOH

1 mol Na2SO4

1

Given over 1

Mole ratio

1 mol = 142.042g

= g Na2SO4

249


If they give you the actual yield and you figure out the theoretical yield, you can find the percent yield.

Actual Yield

Theoretical Yield

X 100

= % Yield

Same as…

% Yield

100

Actual Yield

Theoretical Yield

=


Your percent yield should not be greater than 100 because the theoretical yield is the MAXIMUM yield you can have.


Example: From the examples above, if 221 grams of sodium sulfate were actually collected, what is the percent yield?

From the experiment

221 g NaSO4

249 g NaSO4

X 100

= 88.8%

From the calculation


Practice6
Practice sulfate were actually collected, what is the percent yield?

  • A student conducts a single displacement reaction that produces 2.75 grams of copper. Mathematically he determines that 3.150 grams of copper should have been produced. Calculate the student's percentage yield.

  • A student completely reacts 5.00 g of magnesium with an excess of oxygen to produce magnesium oxide. Analysis reveals 8.10 g of magnesium oxide. What is the student's percentage yield? (*Hint: you must first find the theoretical yield.)

87.3%

97.7%


Practice7
Practice sulfate were actually collected, what is the percent yield?

3. Consider the reaction below of zinc and hydrochloric acid to produce zinc chloride and hydrogen gas.

Zn + 2HCl  H2 +ZnCl2

If a sufficient mass of zinc metal combines with 58.0 grams of HCl, what is the theoretical yield of ZnCl2? What is the actual yield in grams if the percent yield is 87%.

Actual yield: 93.7 g


Practice 1
Practice 1 sulfate were actually collected, what is the percent yield?

  • A student conducts a single displacement reaction that produces 2.75 grams of copper. Mathematically he determines that 3.15 grams of copper should have been produced. Calculate the student's percentage yield.

Actual from the experiment

2.75 g Cu

= 87.3 %

X 100

3.15 g Cu

calculated


Practice 2
Practice 2 sulfate were actually collected, what is the percent yield?

A student completely reacts 5.00g of magnesium with an excess of oxygen to produce magnesium oxide. Analysis reveals 8.10 g of magnesium oxide. What is the student's percentage yield? (*Hint: you must first find the theoretical yield.)

Actual yield

actual

= % yield

X 100

Theo.


To calculate the theoretical yield
To calculate the theoretical yield sulfate were actually collected, what is the percent yield?

2Mg + O2 2 MgO

G:

W:

R:

5.00 g Mg

2 different substances

g MgO

You see g Mg and g MgO

2 mol Mg = 2 mol MgO

1 mol Mg = 24.305 g Mg

1 mol MgO = 40.304 g MgO

5.00 g Mg

1 mol Mg

40.304 g MgO

2 mol MgO

x

x

x

24.305 g Mg

2 mol Mg

1 mol MgO

1

8.29

Theoretical yield= g MgO


Practice 21
Practice 2 sulfate were actually collected, what is the percent yield?

A student completely reacts 5.00g of magnesium with an excess of oxygen to produce magnesium oxide. Analysis reveals 8.10 g of magnesium oxide. What is the student's percentage yield? (*Hint: you must first find the theoretical yield.)

Actual yield

8.10 g MgO

X 100

= 97.7%

8.29 g MgO


Practice 3
Practice 3 sulfate were actually collected, what is the percent yield?

3. Consider the reaction below of zinc and hydrochloric acid to produce zinc chloride and hydrogen gas.

Zn + 2HCl  H2 +ZnCl2

If a sufficient mass of zinc metal combines with 58.0 grams of HCl, what is the theoretical yield of ZnCl2? What is the actual yield in grams if the percent yield is 87%.

Actual yield: 93.7 g


If a sufficient mass of zinc metal combines with 58.0 grams of HCl, what is the theoretical yield of ZnCl2? What is the actual yield in grams if the percent yield is 87%.

Zn + 2HCl  H2 +ZnCl2

G:

W:

R:

58.0 g HCl

2 different substances

g ZnCl2

You see g HCl & g ZnCl2

2 mol HCl = 1 mol ZnCl2

1 mol HCl = 36.461 g HCl

1 mol ZnCl2 = 136.286 g ZnCl2

58.0 g HCl

1 mol HCl

1 mol ZnCl2

136.286 g ZnCl2

x

x

x

36.461 g HCl

2 mol HCl

1 mol ZnCl2

1

108

Theoretical yield= g ZnCl2


Practice 31
Practice 3 of HCl,

  • If a sufficient mass of zinc metal combines with 58.0 grams of HCl, what is the theoretical yield of ZnCl2? What is the actual yield in grams if the percent yield is 87%.

Actual from the experiment

93.7 g

actual

= 87 %

X 100

108 g

calculated


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