iii. stoichiometry stoy – kee – ahm –eh - tree
DESCRIPTION
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. - PowerPoint PPT PresentationTRANSCRIPT
III. StoichiometryStoy – kee – ahm –eh - tree
Chapter 12
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
• 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
o stoikheion, meaning element
o metron, meaning measureo Thus Stoichiometry-
measuring elements!
III. Stoichiometry
oStoichiometry (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
•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
• 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
__H2SO4(aq) + __NaHCO3(s)
__Na2SO4(aq) + __H2O(l) + __CO2(g)
2
2 2
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
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: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: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
For the Reaction:H2SO4 + 2NaHCO3 Na2SO4 + 2H2O +
2CO2
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.)
Skeletal equation:
Balanced equation:
Fe + O2 Fe2O3
4Fe + 3O2 2Fe2O3
Relationships: 4 mol Fe = 3 mol O2
4 mol Fe = 2 mol Fe2O3
3 mol O2 = 2 mol Fe2O3
Practice1. Write three mole ratios (relationships)
from the reaction below:
Al2S3 + H2O Al(OH)3 + H2S1 mol Al2S3 = 6 mol H2O
26 3
1 mol Al2S3 = 2 mol Al(OH)3
1 mol Al2S3 = 3 mol H2S6 mol H2O = 2 mol Al(OH)3
6 mol H2O = 3 mol H2S2 mol Al(OH)3 = 3 mol H2SY
ou s
hou
ld h
ave 3
Practice1. Write three mole ratios (relationships)
from the reaction below:Al2S3 + H2O Al(OH)3 + H2S
1 mol Al2S3 = 6 mol H2O
26 3
1 mol Al2S3 = 2 mol Al(OH)3
1 mol Al2S3 = 3 mol H2S6 mol H2O = 2 mol Al(OH)3
6 mol H2O = 3 mol H2S
Practice2. 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
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:1. Write a balanced reaction.2. Determine your given & want3. 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 1023
of that type• extras like mg or kg, you know what to do
Now you are ready to solve- IT’S GAME TIME.
Game Time:
1. put your GIVEN OVER 12. place your relationships where the
units cancel out diagonally3. everything equal to each other goes
above and below each other4. cancel out your units until you are
left over with your wanted
Here is a flow chart that we will dissect this unit to do
our problems
For most of the examples we will be using this equation:
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
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)
1
4.00 mol H2
3 mol H2
2.67
Given :Want :Relationships:
4.00 mol H2
? mol NH3
2 mol NH3 = 3 mol H2
= mol NH32 mol NH3x
No grams, No molar mass!!
No fun, mc, No 6.02 x 1023!!
= mol N2
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)
1 mol N2
1
7.8 mol NH3
2 mol NH3
3.9
Given :Want :Relationships:
7.8 mol NH3
? mol N2
2 mol NH3 = 1 mol N2
X
No grams, No molar mass!!
No fun, mc, No 6.02 x 1023!!
= moles H2
Ex 3: How many moles of hydrogen react with 13 moles of nitrogen?
Balanced Equation:N2 (g) + 3H2 (g) 2NH3 (g)
1
13 mol N2
1 mol N2
39
Given :Want :Relationships:
13 mol N2
? mol H2
1 mol N2 = 3 mol H2
3 mol H2x
No grams, No molar mass!!
No fun, mc, No 6.02 x 1023!!
2. Moles to Mass/ Mass to Moles
Steps to Success!1. Write a balanced reaction.2. Determine your given & want3. 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
1023 of 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?
Balanced Equation:N2 (g) + 3H2 (g) 2NH3 (g)
Given :Want :Relationships:
4.00 mol H2
? g NH3
2 mol NH3 = 3 mol H2
How do I know when to use mole ratios,
molar mass or 6.02 x 1023?
Look at the “given” and “want” for
clues
1 mol NH3 = 17.031g NH3No fun, mc, No 6.02 x 1023!!
= g NH3
Ex 1: How many grams of ammonia are produced from 4 moles of hydrogen gas in
excess nitrogen?
1
4.00 mol H2
3 mol H2
45.4
2 mol NH3x
Balanced Equation:N2 (g) + 3H2 (g) 2NH3 (g)
Given :Want :Relationships:
4.00 mol H2
? g NH3
2 mol NH3 = 3 mol H21 mol NH3 = 17.031g
NH3
1 mol NH3
17.031g NH3x
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
Remember your Steps
to Success!
1 mol H2 = 2.016g H2
No fun, mc, No 6.02 x 1023!!
= mol NH3
Ex 2: How many moles of ammonia are produced from 4 grams of hydrogen gas in
excess nitrogen?
1
4.00 g H2
2.016g H2
1.32
1 mol H2
x
Balanced Equation:N2 (g) + 3H2 (g) 2NH3 (g)
3 mol H2
2 mol NH3x
Given :Want :Relationships:
4.00 g H2
? mol NH3
2 mol NH3 = 3 mol H2
1 mol H2 = 2.016g H2
Practice1. 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
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
Remember your Steps
to Success!
1 mol H2 = 2.016g H2
1 mol NH3 = 17.031 g NH3
= g NH3
Ex 1: How many grams of ammonia are produced from 4 grams of hydrogen gas in
excess Nitrogen?
1
4.00 g H2
2.016g H2
22.5
1 mol H2x
Balanced Equation:N2 (g) + 3H2 (g) 2NH3 (g)
3 mol H2
2 mol NH3x
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
1 mol NH3
17.031g NH3x
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
Remember your Steps
to Success!
1 mol H2 = 2.016g H2
1 mol N2 = 28.014 g N2
= g N2
Ex 2: How many grams of Nitrogen react with 12.0 grams of Hydrogen?
112.0 g H2
2.016g H2
55.6
1 mol H2x
Balanced Equation:N2 (g) + 3H2 (g) 2NH3 (g)
3 mol H2
1 mol N2x
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
1 mol N2
28.014g N2
x
Practice
1. Determine the mass of NH3 produced from 280 g of N2.
2. What mass of nitrogen is needed to produce 100. kg of ammonia?
340 g NH3
8.22 x 104 g N2
3. MC/F.UN/ Atoms to Mass
Ex 1: How many grams of ammonia are produced from 2.09 x 1014 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 NH31 mol H2 = 6.02 x 1023 mc
H2
Ex 1: How many grams of ammonia are produced from 2.09 x 1014 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
= g NH3
1
2.09 x 1014 mc H2
6.02 x 1023 mc H2
3.94 x 10-9
1 mol H2x3 mol H2
2 mol NH3x1 mol NH3
17.031g NH3x
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
H2 in the presence of excess N2?
= mc NH3
1
40.2g H2
2.016g H2
8.00 x 1024
1 mol H2x3 mol H2
2 mol NH3x1 mol NH3
6.02 x 1023 mc NH3
x
Balanced Equation:N2 (g) + 3H2 (g) 2NH3 (g)
Given :Want :Relationships:
40.2g H2
? mc NH32 mol NH3 = 3 mol H2
1 mol H2 = 2.016g H2
1 mol NH3 = 6.02 x 1023 mc NH3
Practice
1. How many mc of nitrogen will react with exactly 7.04 grams of hydrogen?
2. 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?
= mc N2
1
7.04 g H2
2.016g H2
7.01 x 1023
1 mol H2x3 mol H2
1 mol N2x1 mol N2
6.02 x 1023 mc N2x
Balanced Equation:N2 (g) + 3H2 (g) 2NH3 (g)
Given :Want :Relationships:
7.04 g H2
? mc N21 mol N2 = 3 mol H2
1 mol H2 = 2.016g H2
1 mol N2 = 6.02 x 1023 mc N2
How many mc of ammonia will 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 NH3 = 6.02 x 1023 NH3
1 mol N2 = 6.02 x 1023 mc N2
= mc NH3
1
2.09 x 1013 mc N2
6.02 x 1023 mc N2
4.18 x 1021
1 mol N2x1 mol N2
2 mol NH3x
1 mol NH3
6.02 x 1023 mc NH3x
4. Molar Volume of Gases (Liters)
4. Molar Volume of Gases (liters)
• 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…
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
Remember your Steps
to Success!
1 mol H2 = 22.4 L H2
No grams, No molar mass
No fun, mc, No 6.02 x 1023!!
= mol N2
Ex 1: How many moles of nitrogen gas are needed to react with 44.8 liters of hydrogen
gas to produce ammonia gas?
1
44.8 L H2
22.4 L H2
0.667 1 mol H2
x
Balanced Equation:N2 (g) + 3H2 (g) 2NH3 (g)
3 mol H2
1 mol N2x
Given :Want :Relationships:
44.8 L H2
? mol N2
1 mol N2 = 3 mol H2
1 mol H2 = 22.4 L H2
Given over 1 1 mol = 22.4 L Mole ratio
Ex 2: If 5.00 moles of H2 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
Remember your Steps
to Success!
1 mol NH3 = 22.4 L NH3
No grams, No molar mass
No fun, mc, No 6.02 x 1023!!
= L NH31
5.00 mol H2
3 mol H2
74.7 2 mol NH3x 1 mol NH3
22.4 L NH3x
Given over 1 Mole ratio 1 mol = 22.4 L
Ex 2: If 5.00 moles of H2 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
II. Limiting Reactants and
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?”
• The substance that is not used up completely in a reaction is sometimes called the excess reactant (excess)
• Once one of the reactants is used up, no more product can be formed
“which reactant do you have extra of?”
+ 4
4 13Which is the limiting reactant
4 shells
13 tires
1 shell
4 tires
1 car
1 car = 4 cars can be made if 4 shells are available
= 3.25 cars can be made if 13 tires are available
Less product can be produced, therefore tires must be the limiting reactant
Activity
• 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? 1.21 mol Zn
2.64 mol HCl
Available
2 mol HCl
2 mol HCl1 mol Zn
1 mol Zn
= 2.42 mol HCl
= 1.32 mol Zn
Needed
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
2.64 mol HCl 2.42 mol HCl
<
>
Limiting reactant
Excess reactant
To find the limiting reactant and the excess
reactant:1) Determine the balanced chemical equation2) 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
ExampleSome 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?
Some rocket engines use a mixture of hydrazine, N2H4, and hydrogen peroxide, as the propellant. The
products formed are nitrogen gas and steam.
Step 1: Write a balanced reaction
N2H4 + H2O2 N2(g) + H2O(g)42
Which is the limiting reactant when 0.750 mol of N2H4 is mixed with 0.500 mol of
H2O2?
N2H4 + H2O2 N2(g) + H2O(g)42
0.75 mol N2H4
0.50 mol H2O2
Available
2 mol H2O2
2 mol H2O2
1 mol N2H4
1 mol N2H4
= 1.50 mol H2O2
= 0.25 mol N2H4
Needed
Excess
Limiting
Reactant
To find the limiting reactant and the excess
reactant:4. To calculate the amount of excess:
-Available Amount – Needed Amount = Amount of Excess
How much excess reactant, in moles, remains unchanged?
0.75 mol N2H4
0.50 mol H2O2
Available
2 mol H2O2
2 mol H2O2 =
1 mol N2H4
1 mol N2H4 =
1.50 mol H2O2
0.25 mol N2H4
Needed
Excess
*Available – Needed = Remaining excess
- = 0.50 mol N2H4
To find the limiting reactant and the excess
reactant:5. To calculate the amount of product
produced:-G: amount of available limiting reactant-W: amount of product
How many moles of product are formed?
0.50 mol H2O2
0.50 mol H2O2
G:W:R:
2 mol H2O2
2 mol H2O2 = 1 mol N2
mole N2
1 mol N2 = 0.25 mol N2
N2H4 + H2O2 N2(g) + H2O(g)42G: amount of available limiting reactantN2 is the first product
0.50 mol H2O2
0.50 mol H2O2
G:W:R:
2 mol H2O2
2 mol H2O2 = 4 mol H2Omole H2O
4 mol H2O = 1.0 mol H2O
G: amount of available limiting reactantH2O is the second product
Ex. Zinc metal and solid sulfur (S8) 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?
2.0 mol Zn
1.0 mol S8
Available
1 mol S8
1 mol S8
8 mol Zn
8 mol Zn
= 0.25 mol S8
= 8.0 mol Zn
Needed
Balanced Equation: 8 Zn + S8 8 ZnS
Ex. Zinc metal and solid sulfur (S8) 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
1.0 mol S8 0.25 mol S8
<
>
Limiting reactant
Excess reactant
Ex. Zinc metal and solid sulfur (S8) 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
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
Ex. Zinc metal and solid sulfur (S8) 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?
2 mol Zn
2 mol Zn
G:W:R:
8 mol Zn
8 mol Zn = 8 mol ZnSmol ZnS
8 mol ZnS = 2 mol ZnS formed
How much product was formed?Balanced Equation: 8 Zn + S8 8 ZnS
Practice1. 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?
2.4 mol C
3.1 mol H2O
Available
1 mol H2O
1 mol H2O1 mol C
1 mol C
= 2.4 mol H2O
= 3.1 mol C
Needed
Balanced Equation: C + H2O H2 + CO
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
3.1 mol H2O 2.4 mol H2O
<
>
Limiting reactant
Excess reactant
To find the limiting reactant and the excess
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
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?
2.4 mol CG:W:R: 1 mol C = 1 mol H2
mole H2
G: amount of available limiting reactantH2 is the first product
2.4 mol CG:W:R: 1 mol C = 1 mol CO
mole COG: amount of available limiting reactantCO is the second product
Balanced Equation: C + H2O H2 + CO
2.4 mol C 1 mol H2
1 mol C= 2.4 mol H2
2.4 mol C
1 mol C
1 mol CO = 2.4 mol CO
Set VI: Reactions
2) Zn + Pb(NO3)2 Pb + Zn(NO3)2
3) Fe + 2HCl H2 + FeCl2
III. Theoretical yield, Actual yield and Percent
yield
C. Theoretical yield, Actual yield and Percent yield
• 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 Na2SO4 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 Na2SO4 in grams if 3.50 moles of NaOH is
reacted with sufficient H2SO4?
= g Na2SO4
1
3.50 mol NaOH
2 mol NaOH
249
1 mol Na2SO4
x1 mol Na2SO4
142.042g Na2SO4x
Given over 1 Mole ratio 1 mol = 142.042g
2NaOH + H2SO4 Na2SO4 + 2H2OG:
W:
R:
3.50 mol NaOH? g Na2SO4
2 mol NaOH = 1 mol Na2SO41 mol Na2SO4 = 142.042g Na2SO4
2 different substancesYou see g Na2SO4
If they give you the actual yield and you figure out the
theoretical yield, you can find the percent yield.
Actual YieldTheoretical Yield
X 100 = % Yield
Actual YieldTheoretical Yield
= % Yield 100
Same as…
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?
221 g NaSO4 249 g NaSO4
X 100 = 88.8%
From the experiment
From the calculation
Practice1.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.
2.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%
Practice3. 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 11.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.
2.75 g Cu3.15 g Cu
Actual from the experiment
calculated
X 100 = 87.3 %
Practice 2A 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
Theo.X 100 = % yieldactual
To calculate the theoretical yield
Theoretical yield= g MgO
1
5.00 g Mg 24.305 g Mg
8.29
1 mol Mgx2 mol Mg2 mol MgOx
2Mg + O2 2 MgOG:
W:
R:
5.00 g Mgg MgO
2 mol Mg = 2 mol MgO
1 mol MgO = 40.304 g MgO
2 different substancesYou see g Mg and g MgO
1 mol Mg = 24.305 g Mg
1 mol MgO
40.304 g MgOx
Practice 2A 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
8.29 g MgOX 100 = 97.7%8.10 g MgO
Practice 33. 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%.
Theoretical yield= g ZnCl2
1
58.0 g HCl
36.461 g HCl
108
1 mol HClx
2 mol HCl
1 mol ZnCl2x
Zn + 2HCl H2 +ZnCl2G:
W:
R:
58.0 g HClg ZnCl2
2 mol HCl = 1 mol ZnCl2
1 mol ZnCl2 = 136.286 g ZnCl2
2 different substancesYou see g HCl & g ZnCl2
1 mol HCl = 36.461 g HCl
1 mol ZnCl2
136.286 g ZnCl2
x
actual
Practice 3• 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%.
93.7 g108 g
Actual from the experiment
calculated
X 100 = 87 %