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Slide 1 / 139
StoichiometricCalculations
Slide 2 / 139
Stoichiometry Stoichiometry is the quantitative determination of reactants or products using a balanced chemical equation.
We can interpret a balanced equation in several ways.
The most common way is to interpret it as indicating the number of moles of each substance. For instance, this equation,
3 H2 + N2 ® 2 NH3
can be read as:3 moles of hydrogen gas reacts with 1 mole of nitrogen gas to yield 2 moles of ammonia, or
3 mol H2 + 1 mol N2 to yield 2 mol NH3
The coefficients in the balanced equation give the ratio of moles of reactants and products. Therefore, a balanced chemical equation is needed to perform any stoichiometric calculation.
Slide 3 / 139
Since the coefficients in the balanced equation represent moles of substances, then we can write "mole ratios" which show relative amounts of reactants and products.
Here is an example showing how to write "mole ratios" :
Consider the reaction that produces ammonia, NH3.
N2 + 3 H2 ® 2 NH3
Here are three different mole ratios based on this equation.
1 mol N2
3 mol H2
1 mol N23 mol H2
2 mol NH3 2 mol NH3
Stoichiometric Calculations with Moles
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1 mol N2
3 mol H2
1 mol N23 mol H2
2 mol NH3 2 mol NH3
Stoichiometric Calculations with Moles
The above ratios can be used to determine the quantity of any reactant or product.
N2 + 3 H2 ® 2 NH3
For every 1 mol of N2,· you would need 3 mol of H2 to completely react with it, and· you would produce 2 mol of NH3
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N2 + 3 H2 ® 2 NH3
In fact, there are three more mole ratios that can be written simply by taking the reciprocal of the first three.
1 mol N2
3 mol H2
1 mol N23 mol H2
2 mol NH3 2 mol NH3
1 mol N2
3 mol H2
3 mol H2 1 mol N2
2 mol NH3 2 mol NH3
Stoichiometric Calculations with Moles
It is not important to write down all of the possible mole ratios for a given reaction. But you do need to know how to create a mole ratio for a specific problem.
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Stoichiometric Calculations with Moles
N2 + 3 H2 ® 2 NH3
Sample Problem #1How many moles of NH3 can be made by reacting 4 moles of N2?
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Stoichiometric Calculations with Moles Sample Problem #1How many moles of NH3 can be made by reacting 4 moles of N2?
Sample Problem #1 - Solution· In the problem, circle the "given" quantity and underline the "wanted" quantity.
· Write the "given" quantity. 4 moles of N2 · Select the mole ratio that has the "given" unit in the denominator.
Choose the mole ratio that has the "wanted" unit in the numerator.
1 mol N2
3 mol H2
1 mol N2
2 mol NH3
1 mol N2
3 mol H2
1 mol N2
2 mol NH3
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Stoichiometric Calculations with Moles Sample Problem #1How many moles of NH3 can be made by reacting 4 moles of N2?
Sample Problem #1 - Solution (con't)
We are now ready to set up our problem: 4 moles of N2
Just as you multiply fractions, multiply the numbers across the top and divide by any numbers in the denominators.
1 mol N2
2 mol NH3x
1
1 mol N2
2 mol NH3x
1
4 moles of N2= 8 mol NH3
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N2 + 3 H2 ® 2 NH3
Sample Problem #2How many moles of H2 will be needed to react with 15 mol N2?
Stoichiometry Calculations with Moles
The same approach can tell you how many moles of one reactant is needed to completely react with another reactant.
1 mol N2
3 mol H2
1 mol N23 mol H2
2 mol NH3 2 mol NH3
1 mol N2
3 mol H2
3 mol H2 1 mol N2
2 mol NH3 2 mol NH3
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1 mol N2
3 mol H2
1 mol N23 mol H2
2 mol NH3 2 mol NH3
1 mol N2
3 mol H2
3 mol H2 1 mol N2
2 mol NH3 2 mol NH3
Stoichiometric Calculations with Moles
Sample Problem #2- Solution
How many moles of H2 will be needed to react with 15 mol N2?
15 moles of N2 3 mol H2
1 mol N21x = 45 mol H2
We choose the mole ratio that has mol N2 in the denominator and mol H2 in the numerator.
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= 4 mol Al2O3
= 6 mol O2 x 2 molAl2O3 3 mol O2
1 What is the largest number of moles of Al2O3 that could result from reacting 6 moles of O2?
4 Al (s) + 3 O2 (g) ® 2 Al2O3 (s)
Answer
A 2 mol
B 3 mol
C 4 mol
D 18 mol
E 24 mol
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= 18 mol O2
2 How many moles of O2 would be required to create 12 moles of Al2O3?
4 Al (s) + 3 O2 (g) ® 2 Al2O3 (s)
= 12 mol Al2O3 x 3 mol O2 2 mol Al2O3
Answer
A 3 mol
B 4 mol
C 8 mol
D 18 mol
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8mol Al x 3 mol O2 ----------- 4 mol Al
= 6 mol O2
3 How many moles of O2 would be required to completely react with 8 moles of Al?
4 Al (s) + 3 O2 (g) ® 2 Al2O3 (s)
Answer
A 2 mol
B 3 mol
C 6 mol
D 12 mol
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4 When iron rusts in air, iron (III) oxide is produced. How many moles of oxygen react with 2.4 mol Fe?
4 Fe (s) + 3 O2 (g) ® 2 Fe2O3 (s)
2.4 mol Fe x 3 mol O2 ----------- 4 mol Fe
= 1.8 mol O2
Answer
A 1.2 mol
B 1.8 mol
C 2.4 mol
D 3.2 mol
E 4.8 mol
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5 How many moles of Al are needed to react completely with 1.2 mol FeO?
1.2 mol FeO x 2 mol Al ----------- 3 mol FeO
= 0.8 mol Al
Answer
2 Al (s) + 3 FeO (g) ® 3 Fe (s) + Al2O3 (s)
A 0.8 mol
B 1.2 mol
C 1.6 mol
D 2.4 mol
E 4.8 mol
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6 How many moles of calcium metal are produced from the decomposition of 8 mol of calcium chloride?
CaCl2 (s) ® Ca (s) + Cl2 (g)
8 mol CaCl2 x 1 mol Ca ---------------- 1 mol CaCl2
= 8 mol Ca
Answer
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8 mol Ca8 mol Cl2Total of 16 moles
7 How many moles (total) of calcium metal and chlorine gas are produced from the decomposition of 8 mol of cal cium chloride?
CaCl2 (s) ® Ca (s) + Cl2 (g)
Answer
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Silver and nitric acid react according to the following balanced equation:
3 Ag(s) + 4 HNO3(aq) à 3 AgNO3(aq) + 2 H2O(l) + NO(g)
A. How many moles of silver are needed to react with 40 moles of nitric acid, HNO3?
B. Using 40 mol HNO3, how many moles of silver nitrate will be produced?
C. Using 40 mol HNO3, how many moles of water will be produced?
D. Using 40 mol HNO3, how many moles of nitrogen monoxide will be made?
Stoichiometric Calculations with Moles
Slide 19 / 139
Answers:
A) 30 mol Ag
B) 30 mol AgNO3
C) 20 mol H2O
D) 10 mol NO
Slide 20 / 139
2 N2H4(l) + N2O4(l) à 3 N2(g) + 4 H2O(g)
A. How many moles of dinitrogen tetrahydride are required to produce 57 moles of nitrogen?
B. How many moles of dinitrogen tetroxide are required to produce 57 moles of nitrogen?
C. How many moles of water are produced when 57 moles of nitrogen are made?
Stoichiometric Calculations with Moles
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Answers:
A) 38 mol N2H4
B) 19 mol N2O4
C) 76 mol H2O
Slide 22 / 139
Stoichiometry
So far, we have interpreted the coefficients in a balanced equation as indicating the number of moles of each substance. However, we said before that there are many different ways to interpret a balanced equation.
The coefficients in a balanced equation can represent· moles· representative particles (atoms, molecules, formula units)· liters of gas (at constant temperature & pressure)
The chart on the next slide shows these relationships.
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3 H2 N2 2 NH3
3 moles of H2 1 mole of N2 2 moles of NH3
3 molecules of hydrogen (H2)
1 molecule of nitrogen (N2)
2 molecules of ammonia (NH3)
3 L of H2 * 1 L of N2 * 2 L of NH3 *
* at same Temperature & Pressure
Interpreting a Balanced Equation
3H2 + N2 à 2NH3
· moles· representative particles (atoms, molecules, formula units)· liters of gas *
The coefficients in a balanced equation can represent
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Interpreting a Balanced Equation
3H2 + N2 à 2NH3
· In terms of moles, the equation above showsa total of 4 moles reacting to yield 2 moles of product.
· In terms of particles, a total of 4 molecules react to form 2 molecues.
· In terms of liters, a total of 4 liters of gas react to form 2 liters of gas (at STP, 89.2 L of gas react to yield 44.8 L).
The number of moles, particles and liters before the arrow is not the same as after the arrow. Therefore, these quantities are not conserved in a balanced equation.
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Interpreting a Balanced Equation
3H2 + N2 à 2NH3
We have seen various ways to interpret a balanced equation.
Recall that a balanced equation is one in which both sides have the same number of each kind of atom. As a result, the Law of Conservation of Mass is obeyed, since no atoms can be created nor destroyed.
Therefore, the quantities that are conserved in a balanced equation are mass and atoms. Conserved means that the amount that appears before the arrow is the same as the amount after the arrow.
Only mass and the number of atoms are conserved in a balanced equation.
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3 H2 N2 2 NH3
6 atoms of hydrogen (H)
2 atoms of nitrogen (N)
6 atoms of hydrogen (H)and
2 atoms of nitrogen (N)
3 mol x 2 g/mol
= 6 g of H2
1 mol x 28 g/mol
= 28 g of N2
2 mol x 17 g/mol
= 34 g of NH3
Interpreting a Balanced Equation
3H2 + N2 à 2NH3
Only mass and the number of atoms are conserved in a balanced equation.
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8 In a chemical reaction that is represented by a balanced equation, the quantities that are conserved are
A mass and molecules
B mass and atoms
C moles and liters
D moles and molecules
E mass and liters
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9 In a chemical reaction that is represented by a balanced equation, the quantities that are NOT conserved are
A mass and moles
B mass and atoms
C moles and particles
D mass and atoms
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Stoichiometry Calculations with ParticlesWe will now consider a second interpretation of a balanced equation in terms of particles (which could be molecules, atoms, or formula units).
Multiplying the number of moles by 6.02 x 1023 yields the numbers of particles. So a formula that's balanced for moles must also be balanced for particles.
3 H2 + N2 ® 2 NH3can be read as:
3 molecules of H2 plus 1 molecule of N2 yields 2 molecules of NH3
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Stoichiometry Calculations with Particles
Solving stoichiometry problems with particles is similar to those with moles. One big difference is that your answers for particles cannot be fractions or decimal numbers; they must be whole numbers.
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Stoichiometry Calculations with ParticlesSo here are the six ratios that you could use for solving problems involving particles:
3 H2 + N2 ® 2 NH3
2 molecules NH3
3 molecules H2
3 molecules H2
3 molecules H2
1 molecule N2
1 molecule N2
1 molecule N2
2 molecules NH3
2 molecules NH3
and their reciprocals
3 molecules H2
2 molecules NH3
1 molecule N2
Note that there is no abbreviation for "molecule."
Slide 32 / 139
Stoichiometry Calculations with Particles
Sample Problem #3How many molecules of N2 are needed to completely react with 24 molecules of H2?
Sample Problem #3 - Solution · In the problem, circle the "given" quantity and underline the "wanted" quantity.
How many molecules of N2 are needed to completely react with 24 molecules of H2?
· Write the "given" quantity.
24 molecules of H2 1
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Stoichiometry Calculations with ParticlesSample Problem #3How many molecules of N2 are needed to completely react with 24 moles of H2?
Sample Problem #3 - Solution (con't)· Select the ratio that has the "given" unit in the denominator and the "wanted" unit in the numerator.
2 molecules NH3
3 molecules H2
3 molecules H2
3 molecules H2
1 molecule N2
1 molecule N2
1 molecule N2
2 molecules NH3
2 molecules NH3
3 molecules H2
2 molecules NH3
1 molecule N2
24 molecules of H2 1
Slide 34 / 139
Stoichiometry Calculations with ParticlesSample Problem #3How many molecules of N2 are needed to completely react with 24 moles of H2?
Sample Problem #3 - Solution (con't)· Select the ratio that has the "given" unit in the denominator and the "wanted" unit in the numerator.
24 molecules of H2 1 3 molecules H2
1 molecule N2x = 8 molecules N2
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10 What is the largest number of of Li3N formula units that could result from reacting 6 N2 molecules?
6 Li (s) + N2 (g) ® 2 Li3N (s)
6 molecules N2 x 2 formula units Li3N ---------------------- 1 molecule N2
= 12 formula units Li3N
Answer
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11 How many N2 molecules would be required to create 4 Li3N formula units?
6 Li (s) + N2 (g) ® 2 Li3N (s)
4 formula units Li3N x 1 molecule N2 ---------------------- 2 formula units Li3N
= 2 molecules N2
Answer
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12 How many Li atoms would be required to completely react with 3 N2 molecules?
6 Li (s) + N2 (g) ® 2 Li3N (s)
3 molecules N2 x 6 atoms Li ---------------------- 1 molecule N2
= 18 atoms Li
Answer
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Stoichiometry Calculations with Volume
So far, we have seen that coefficients in a balanced equation can represent moles or representative particles. A third interpretation applies only to gases involved. If all the gases are at the same temperature and pressure, then the coefficients give the relative volumes of gas in liters.
CS2 (g) + 3 O2 (g)® CO2 (g) + 2 SO2 (g)
This formula can be read as:*
1 L of CS2 + 3 L of O2 yields 1 L of CO2 + 2 L of SO2
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Stoichiometry Calculations with VolumeUsing coefficients to represent volume only works when comparing two gases in a formula. It cannot be used to compare a gas to a liquid or solid.
CS2 (g) + 3 O2 (g)® CO2 (g) + 2 SO2 (g)
At STP conditions:
1 x 22.4 L + 3 x 22.4 L ® 1 x 22.4 L + 2 x 22.4 L
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Using this interpretation we can directly answer questions about the relative volumes of gas reactants and products.
For the next problem, we will not write out all of the ratios ahead of time.
Sample Problem #4How many liters of SO2 will be produced when 6 liters of O2 are reacted as shown below?
CS2 (g) + 3 O2 (g)® CO2 (g) + 2 SO2 (g)
Stoichiometry Calculations with Gases
Slide 41 / 139
Sample Problem #4 - Solution
How many liters of SO2 will be produced when 6 liters of O2 are reacted as shown below?
CS2 (g) + 3 O2 (g)® CO2 (g) + 2 SO2 (g)
Stoichiometry Calculations with Gases
· Write the "given" quantity.· Create a ratio that has the "given" unit in the denominator and the "wanted" unit in the numerator.
6 liters of O2
3 L of O2
2 L of SO2x = 4 L of SO2
Slide 42 / 139
13 The coefficients in a balanced equation can represent
A mass, in grams
B volume, in liters (gases only)
C representative particles
D Both A and B
E Both B and C
Slide 43 / 139
= 8 L of H2O
14 How many liters of H2O (g) will be created from reacting 8.0 L of H2 (g) with a sufficient amount of O2 (g)?
2 H2 (g) + O2 (g) ® 2 H2O (g)
Answer
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15 The equation below shows the decomposition of solid lead (II) nitrate, Pb(NO3)2. How many liters of O2 will be produced when 12 L of NO2 are formed?
2 Pb(NO3)2 (s) ® 2 PbO (s) + 4 NO2 (g) + O2 (g)
= 3 L of O2
= 12 L NO2 x 1 L O2 4 L NO2
Answer
A 1.0 L
B 2.0 L
C 3.0 L
D 4.0 L
E 12 L
Slide 45 / 139
= 250 L of O2
16 What volume of methane is needed to completely react with 500 L of O2 at STP? You must first balance the equation.
__ CH4 + ___ O2 --> ___ CO2 + ___ H2O
= 500L O2 x 1 L CH4 2 L O2
Answer
A 125 L
B 250 L
C 500 L
D 1000 L
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= 36 L O2 x 4 L NO2 7 L O2
= 21 L of NO2
17 How many liters of NO2 (g) will be created from reacting 36 L of O2 (g) with a sufficient amount of NH3 (g)?
4 NH3 (g) + 7 O2 (g) ® 4 NO2 (g) + 6 H2O (g)
Answer
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So far, we have seen stoichiometry problems that can be solved in one step, using only coefficients. These are: · mole-mole problems, · particle-particle problems and · volume-volume problems.
Coefficients can represent moles, particles, or liters.
When mass is involved, however, we must now take into consideration the fact that all particles have a different molar mass.
Coefficients do not represent mass in grams.
Stoichiometric Calculations with Mass
Slide 48 / 139
Stoichiometric Calculations with Mass
3 H2 N2 2 NH3
3 moles of H2 1 mole of N2 2 moles of NH3
3 mol x 2 g/mol
= 6g of H2
1 mol x 28 g/mol
= 28g of N2
2 mol x 17 g/mol
= 34g of NH3
3 H2 + N2 ® 2 NH3
Using this example once again, it is clear that 3 grams of H2 + 1 gram of N2 will not equal 2 grams of NH3.
Coefficients do not represent mass in grams.
6 g of H2 + 28 g of N2 will equal 34 g of NH3
Slide 49 / 139
Stoichiometric Calculations with Mass
Consider this question asked earlier:What is the largest number of moles of Al2O3 that could result from reacting 6 moles of O2?
4 Al (s) + 3 O2 (g) --> 2 Al2O3 (s)
Now, instead of asking for moles of Al2O3 , consider this question:
Sample Problem #5What is the mass, in grams, of Al2O3 that could result from reacting 6 moles of O2?
Slide 50 / 139
Sample Problem #5 - Solution
What is the mass, in grams, of Al2O3 that could result from reacting
6 moles of O2?
We proceed similarly as before, changing from moles of the "given" quantity to the moles of the "wanted" quantity.
Stoichiometric Calculations with Mass
4 Al (s) + 3 O2 (g) --> 2 Al2O3 (s)
6 moles of O2
3 mol O2
2 mol Al2O3x1
= 4 mol Al2O3
This answer becomes the new "given" which we simply convert to grams, using the molar mass of Al2O3.
Slide 51 / 139
Sample Problem #5 - Solution (con't)
What is the mass, in grams, of Al2O3 that could result from reacting 6 moles of O2?
Stoichiometric Calculations with Mass
4 Al (s) + 3 O2 (g) --> 2 Al2O3 (s)
4 mol Al2O3
1 mol Al2O3
102 g Al2O3x1
= 408 g Al2O3
Remember: Do not use coefficients when converting between moles and mass. The molar mass is for only ONE mole!
Slide 52 / 139
Stoichiometric Calculations with Mass
An alternative way to solve the previous problem is to set-up both ratios, then perform the calculation without stopping after the first step.
6 moles of O2
3 mol O2
2 mol Al2O3x1 1 mol Al2O3
102 g Al2O3x = 408 g Al2O3
The key is to make sure that the unit in any numerator appears in the next denominator so that units will cancel out.
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= 18 mol O2 x 32 g O2 1 mol O2
= 576 g O2
18 What mass, in grams, of O2 would be required to create 12 moles of Al2O3?
4 Al (s) + 3 O2 (g) ® 2 Al2O3 (s)
= 12 mol Al2O3 x 3 mol O2 2 mol Al2O3
Answer
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= 192 g O2
8mol Al x 3 mol O2 ----------- 4 mol Al
= 6 mol O2 x 32 g O2 1 mol O2
Answer
19 How many grams of O2 would be required to completely react with 8 moles of Al?
4 Al (s) + 3 O2 (g) ® 2 Al2O3 (s)
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20 When iron rusts in air, iron (III) oxide is produced. How many grams of oxygen react with 2.4 mol Fe?
4 Fe (s) + 3 O2 (g) ® 2 Fe2O3 (s)
2.4 mol Fe x 3 mol O2 ----------- 4 mol Fe
= 1.8 mol O2 x 32 g O2 1 mol O2
= 57.6 g O2
Answer
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21 How many grams of Al are needed to react completely with 1.2 mol FeO?
1.2 mol FeO x 2 mol Al ----------- 3 mol FeO
= 0.8 mol Al x 27 g Al 1 mol Al
= 21.6 g Al
Answer
2 Al (s) + 3 FeO (g) ® 3 Fe (s) + Al2O3 (s)
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Mass-Mass Calculations
Given Find
We started this unit by using coefficients in a balanced equation to solve mole-mole problems.
Slide 58 / 139
Next, we extended these problems by calculating the mass of the "wanted" quantity.
Mass-Mass Calculations
Given
Find
Slide 59 / 139
Now, we will solve problems when both the "given" quantity and the "wanted" quantity are in grams. Recall that you cannot simply use the coefficients to relate mass in grams.
Mass-Mass Calculations
Grams of substance A Grams of
substance B
Moles of substance A
Moles of substance B
Use molar mass of A Use molar
mass of B
Use coefficients of A and B
from balanced equation
GivenFind
Slide 60 / 139
A typical "mass-mass" problem is has three steps as outlined below. Remember that coefficients are used ONLY between moles.
Mass-Mass Calculations
Grams of substance A Grams of
substance B
Moles of substance A
Moles of substance B
Use molar mass of A Use molar
mass of B
Use coefficients of A and B
from balanced equation
GivenFind
Step 1
Step 2
Step 3
Slide 61 / 139
Mass-Mass Calculations
Sample Problem #6Calculate the mass of ammonia, NH3, produced by the reaction of 5.4 g hydrogen gas with an excess of nitrogen.
N2 + 3H2 ---> 2NH3
Sample Problem #6 - SolutionCalculate the mass of ammonia, NH3, produced by the reaction of 5.4 g hydrogen gas with an excess of nitrogen.
We will proceed using these steps:
a) convert 5.4 g H2 to moles H2
b) convert moles H2 to moles NH3
c) convert moles NH3 to grams NH3
Slide 62 / 139
a) convert 5.4 g H2 to moles H2
c) convert moles NH3 to grams NH3
b) convert moles H2 to moles NH3
Mass-Mass Calculations
gH2 --> mol H2 --> mol NH3 --> g NH3
5.40 g H2 1 mol H22.0 g H2
17.0 g NH31 mol NH3
2 mol NH33 mol H2
X = 31 g NH3X X
Sample Problem #6 - Solution (con't)
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Sample Problem #7How many grams of water can be produced from the combustion of 1.00 g of glucose?
C6H12O6 + 6 O2 --> 6 CO2 + 6 H2O
Sample Problem #7 - Solution
Our general strategy isStep (a): convert mass of glucose to moles of glucose· use the molar mass of glucose
Step (b): convert moles glucose to moles water· use the ratio of coefficients in the balanced equation
Step (c): convert moles of water to grams of water· use the molar mass of water
Mass-Mass Calculations
Slide 64 / 139
C6H12O6 + 6 O2 --> 6 CO2 + 6 H2O
Mass-Mass Calculations
1mol glucose180.0 g glucose
5.56 x 10-3 mol glucose 6 mol water
1 mol glucose3.33 x 10-2 mol water
18.0 g water
1 mol water
0.600 g water1.00 g glucose
X
X
X
(a) (c)
no direct calculation
(b) for every 1 mol glucoseyou get 6 mols of water
Sample Problem #7 - Solution (con't)How many grams of water can be produced from the combustion of 1.00 g of glucose?
Slide 65 / 139
Mass-Mass Calculations
1.00 g glucose
11mol glucose
180.0 g glucosex x 6 mol water
1 mol glucose
18.0 g water
1 mol waterx
Grams of substance A Grams of
substance B
Moles of substance A
Moles of substance B
Use molar mass of A Use molar
mass of B
Use coefficients of A and B
from balanced equation
GivenFind
= 0.600 g water
Again, the key to this method is making sure that the unit in any numerator appears in the next
denominator so that units will cancel out.
Slide 66 / 139
22 What mass of Na is produced by the decomposition of 6.5 g NaN3?
A 2.3 g
B 4.6 g
C 6.5 g
D 23 g
E 46 g
2 NaN3 (s) ® 2 Na (s) + 3 N2 (g)
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23 How many grams of Al2O3 will be created from reacting 54 g of Al with a sufficient amount of O2?
4 Al (s) + 3 O2 (g) ® 2 Al2O3 (s)
A 27 g
B 54 g
C 102 g
D 204 g
Slide 68 / 139
24 How many grams of Mg must react in order to to create 160 g of MgO?
2 Mg (s) + O2 (g) ® 2 MgO (s)
A 48 g
B 80 g
C 96 g
D 160 g
Slide 69 / 139
25 Approximately how many grams of oxygen are needed to react with 240 g of Mg?
2 Mg (s) + O2 (g) ® 2 MgO (s)
A 80 g
B 120 g
C 160 g
D 240 g
Slide 70 / 139
Mixed Stoichiometry CalculationsThus far, our problem-solving has focused on one of four main types. Most practical applications of chemistry are not this narrowly defined. Many problems in advanced chemistry give a quantity in one unit (e.g. moles, grams, or liters) and ask for the proportional quantity in a different unit.
For example, this problem gives a quantity in moles, but asks for mass, in grams.
How many grams of ammonia can be produced by using 10 mol of nitrogen gas and an unlimited or excess amount of hydrogen gas?
Slide 71 / 139
Mixed Stoichiometry CalculationsEvery type of stoichiometry calculation may be solved by following this map.
(1) From left to right, we convert any "Given" substance to moles. (2)Next, using the mole ratio created with coefficients, one can calculate
the moles of the "Wanted" quantity. (3) Finally, if necessary, moles can be converted to either
particles, mass or volume.(1)
(2)
(3)1 mol G
6.02 x10-23
represenative particles of G X
1 mol G
mass G
mass of G X
1 mol G
22.4 L Gvolume of Gat STP
X
mass W
1 mol W
massof W=
22.4 L W
1 mol WVolume of W at STP=
b mol W
a mol Gmol G X mol W
x
x
6.02 x 1023
1 mol W
representative particles of W=x
Slide 72 / 139
N2 + 3H2 ---> 2NH3
Sample Problem #8How many grams of ammonia can be produced by using10 mol of nitrogen gas and an unlimited or excess amount of hydrogen gas?
Sample Problem #8 - Solution
Since the "Given" quantity is already in moles, we can skip to step (2).
Overview:mol N2 ---> mol NH3 ---> grams NH3
Mixed Stoichiometry Calculations
Slide 73 / 139
10 mol N2 2 mol NH3
1 mol N2
x = 340 g NH3x1 mol NH3
17 g NH3
N2 + 3H2 ---> 2NH3
Sample Problem #8 - Solution (con't)
Mixed Stoichiometry Calculations
Remember: Do not use coefficients when using molar mass.
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Sample Problem #9The airbag in a car generates a relatively large amount of gas quickly through the following reaction.
2 NaN3 (s) ® 2 Na (s) + 3 N2 (g)
How many grams of NaN3 will be required to generate enough gas to fill an airbag with a volume of 36 L N2 (assume STP conditions)?
Sample Problem #9 - Solution
How many grams of NaN3 will be required to generate enough gas to fill an airbag with a volume of 36 L N2 (assume STP conditions)?
Mixed Stoichiometry Calculations
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1 mol G
6.02 x10-23
represenative particles of GX
1 mol G
mass G
mass of G X
1 mol G
22.4 L Gvolume of Gat STP
X
mass W
1 mol W
massof W=
22.4 L W
1 mol WVolume of W at STP=
b mol W
a mol Gmol G X mol W
x
x
6.02 x 1023
1 mol W
representative particles of W=x
(1)
(2) (3)
2 NaN3 (s) ® 2 Na (s) + 3 N2 (g)
Sample Problem #9 - Solution OverviewStep (1) - Liters of N2 --> Moles of N2Step (2) - Moles of N2 --> Moles of NaN3Step (3) - Moles of NaN3 --> Grams of NaN3
Mixed Stoichiometry Calculations
Remember that coefficients are used ONLY between moles.
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2 NaN3 (s) ® 2 Na (s) + 3 N2 (g)
Sample Problem #9 - Solution Step (1) - Liters of N2 --> Moles of N2
Mixed Stoichiometry Calculations
Step (2) - Moles of N2 --> Moles of NaN3
Step (3) - Moles of NaN3 --> Grams of NaN3
36 L N2 1 mol N2
22.4 L N2x = 1.6 mol N2
1
2 mol NaN3
3 mol N2x
1.6 mol N2
1 = 1.1 mol NaN3
65 g NaN3
1 mol NaN3x
1.1 mol NaN3
1 = 72 g NaN3
Remember that coefficients are used ONLY between moles.
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26 How many liters of O2 (g) at STP are required to create 102 g of Al2O3 (s)?
4 Al (s) + 3O2 (g) ® 2 Al2O3 (s)
x x2 mol Al2O3
3 mol O2 102 g Al2O3 1 mol Al2O3 102 g Al2O3 1 mol O2
22.4 L O2 x1
= 33.6 L O2
Answer
A 11.2 L
B 22.4 L
C 33.6 L
D 44.8 L
E 67.2 L
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27 How many grams of Cl2 are needed to react with 1 mol of antimony, Sb?
2 Sb + 3 Cl2 à 2 SbCl3
A 71 g
B 107 g
C 142 g
D 213 g
x x1 mol Sb 3 mol Cl2 2 mol Sb 1 mol Cl2
71 g Cl2
1= 107 g Cl2
Answer
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A. How many manganese atoms are produced if 55 moles of MnO2 react with excess aluminum.
B. How many moles of aluminum oxide are made if 3580 g of manganese oxide are consumed?
3MnO2(s) + 4 Al(s) à 2 Al2O3(s) + 3Mn(s)
*Mixed Stoichiometry Practice
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3MnO2(s) + 4 Al(s) à 2 Al2O3(s) + 3Mn(s)
*Mixed Stoichiometry Practice
C. How many moles of manganese oxide will react with 5.33 x 1025 atoms of aluminum?
D. If 4.37 moles of aluminum are consumed, how many formula units of aluminum oxide are produced?
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Answers:
A) 3.3 x 1025 atoms Mn
B) 27.5 mol Al2O3
C) 66.4 mol MnO2
D) 1.32 x 1024 formula units Al2O3
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The compound tristearin (C57H110O6) is a type of fat which camels store in their hump. Besides being a source of energy, the fat is a source of water for the camel because when the fat is burned, the following combustion reaction occurs:
2 C57H110O6(s) + 163 O2(g) à 114 CO2(g) + 110 H2O(l)
A. At STP, what volume of O2 is required to consume 0.64 moles of tristearin?
B. At STP, what volume of carbon dioxide is produced in Part A?
**Mixed Stoichiometry Practice
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2 C57H110O6(s) + 163 O2(g) à 114 CO2(g) + 110 H2O(l)
C. If 22.4 L of oxygen is consumed at STP, how many moles of water are produced?
D. Find the mass of tristearin required to produce 55.56 moles of water.
**Mixed Stoichiometry Practice
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Answers:
A) 1168 L O2 (or 1200 L)
B) 817 L CO2 (or 820 L)
C) 0.00414 mol H2O
D) 899 g C57H110O6
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Limiting Reagent & Excess Reagent
Percent Yield & Theoretical Yield
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How Many Cookies Can I Make?
· You can make cookies until you run out of one of the ingredients.· Once this family runs out of sugar, they will stop making cookies (at least any cookies you would want to eat).
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In this example the egg would be the limiting reactant, because it will limit the amount of cookies you can make.
How Many Cookies Can I Make?
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How Many Cookies Can I Make?
Recipe for Butter Cream Cookies
1 c. butter 1 c. sugar 2 T. vanilla
1-3oz. pkg cream cheese
1 egg yolk 2-1/2 c. sifted cake flour
This recipe will make 5 dozen cookies.
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28 How many dozen cookies could you make with 3 c. butter and enough of all the other ingredients?
A 1 doz
B 3 doz
C 5 doz
D 10 doz
E 15 doz
Recipe for Butter Cream Cookies
1 c. butter 1 c. sugar 2 T. vanilla
1-3oz. pkg cream cheese
1 egg yolk 2-1/2 c. sifted cake flour
Yields 5 dozen cookies.
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29 How many dozen cookies could you make with 10 c. sifted cake flour and enough of all the other ingredients?
A 2 doz
B 5 doz
C 10 doz
D 20 doz
E 40 doz
Recipe for Butter Cream Cookies
1 c. butter 1 c. sugar 2 T. vanilla
1-3oz. pkg cream cheese
1 egg yolk 2-1/2 c. sifted cake flour
Yields 5 dozen cookies.
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30 How many dozen cookies could you make with 1 T. vanilla and enough of all the other ingredients?
A 1 doz
B 2 doz
C 2-1/2 doz
D 5 doz
E 10 doz
Recipe for Butter Cream Cookies
1 c. butter 1 c. sugar 2 T. vanilla
1-3oz. pkg cream cheese
1 egg yolk 2-1/2 c. sifted cake flour
Yields 5 dozen cookies.
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Limiting Reagents & Theoretical Yield
· The limiting reactant, or limiting reagent, is the reactant present in the smallest stoichiometric amount.
· This is not necessarily the one with the smallest mass.
· The limiting reactant is the reactant you’ll run out of first, and it is the one that determines the maximum amount of product that can be made.
· Theoretical yield is the the maximum amount of product that can be made, based on the limiting reagent.
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Recipe for Butter Cream Cookies
1 c. butter 1 c. sugar 2 T. vanilla
1-3oz. pkg cream cheese
1 egg yolk 2-1/2 c. sifted cake flour
Yields 5 dozen cookies.
Limiting Reagents
In the three previous Senteo questions, the amount given in the problem was the limiting reagent.
· 3 c butter· 10 c. sifted cake flour· 1T. vanilla
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Limiting Reagents
Limiting reagent Excess reagent Theoretical yield
3 c. butter all other ingredients 15 doz.
10 c. sifted cake flour
all other ingredients 20 doz.
1 T. vanilla all other ingredients 2-1/2 doz.
Note that in every case, the theoretical yield is determined by the limiting reagent, not the excess reagents.
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31 When two substances react, the one that is used up first is called the:
A determining reagent
B limiting reagent
C unlimited reagent
D excess reagent
E reactive reagent
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Limiting reagent problems are different from those done previously in that two quantities are given, instead of just one.
It is your job to figure out which reactant is limiting because that will determine the maximum yield.
There are a variety of methods one can use to determine which reactant is the limiting one. Once you have identified the limiting reagent, then the other reactant is the excess reagent.
Limiting Reagents
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Definitions· Limiting reagent - any reactant that is used up firsts in a chemical reaction; it determines the amount of product that can be formed in the reaction
· Excess reagent - a reagent present in a quantity that is more than sufficient to react with a limiting reagent; any reactant that remains after the limiting reagent is used up in a chemical reaction
· Theoretical yield - the maximum amount of product that can be made, based on the limiting reagent
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This is how you identify which is the limiting reagent:
1. Write a balanced equation.
2. Use the given amount of each reactant to calculate the amount of product that could be formed. Use the problem-solving methods for stoichiometry that you learned earlier in this unit.
3. Compare the two amounts of product that could be made. The smaller of the two amounts indicates the maximum amount of product that could be made and is called theoretical yield. The larger of the two amounts is irrelevant and meaningless.
4. Whichever reactant yields the smaller amount of product is thus the limiting reagent.
Limiting Reagent & Excess Reagent
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Sample Problem #10Consider the reaction between hydrogen and oxygen to yield water. 2 H2 (l) + O2 (g) --> 2 H2O (l)
Starting with 10 molecules of H2 and 7 molecules of O2, which reactant will run out first?
Limiting Reagent & Excess ReagentO2 H2
Sample Problem #10 - SolutionUse stoichiometry methods to calculate the following:
10 molecules of H2 will yield ___ molecules of H2O
7 molecules of O2 will yield ___ molecules of H2O
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Sample Problem #10 - Solution (con't)
10 molecules H2
10 molecules of H2 will yield __10_ molecules of H2O
7 molecules O2
7 molecules of O2 will yield __14_ molecules of H2O
Limiting Reagent & Excess Reagent2 H2 (l) + O2 (g) --> 2 H2O (l)
2 molecules H2 x
x
2 molecules H2O
2 molecules H2O
1 molecule O2
Now, we compare the two amounts.
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Sample Problem #10 - Solution (con't)
10 molecules of H2 will yield __10_ molecules of H2O
7 molecules of O2 will yield __14_ molecules of H2O
· The smaller amount is the theoretical yield: in reality, 10 molecules of water can be produced.
· 10 molecules of H2 is the limiting reagent
· 7 molecules of O2 is the excess reagent.
· The amount of 14 molecules of water cannot in fact be made; this would require 14 molecules of H2 which is not available. The amount of 14 molecules is meaningless and serves only to compare to the 10 molecules of water that can be produced.
Limiting Reagent & Excess Reagent
2 H2 (l) + O2 (g) --> 2 H2O (l)
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Sample Problem #11What mass of ammonia can be produced from 65 g nitrogen and 25 g hydrogen?
Sample Problem #11 - SolutionThis problem asks for theoretical yield; however, we must first determine which reagent is limiting.1. Write a balanced chemical equation. N2 + 3H2 ---> 2NH3
2. Calculate how much product can be made from the first "given" amount, 65 g nitrogen.
3. Calculate how much product can be made from the second "given" amount, 25 g hydrogen.
Limiting Reagent & Excess Reagent
x x =65 g N2
28 g N2
1 mol N2
1 1 mol N2
2 mol NH3
1 mol NH3
17 g NH3x 79 g NH3
x x =25 g H2
2 g H2
1 mol H2
1 3 mol H2
2 mol NH3
1 mol NH3
17 g NH3x 142 g NH3
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Sample Problem #11What mass of ammonia can be produced from 65 g nitrogen and 25 g hydrogen?
Sample Problem #11 - Solution (con't)4. Compare the two amounts. The smaller of the two amounts is the answer to the problem. This is called the theoretical yield.
Answer: 79 g of NH3 can be produced; nitrogen is the limiting reagent and hydrogen is the excess reagent.
Limiting Reagent & Excess Reagent
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32 If you are provided with 75 molecules of H2 and 50 molecules of N2, what are the limiting reagent & theoretical yield for the reaction:3 H2 (g) + N2 (g) --> 2 NH3 (g)
A 75 molecules H2; 25 molecules NH3
B 75 molecules H2; 50 molecules NH3
C 50 molecules N2; 50 molecules NH3
D 50 molecules N2; 100 molecules NH3
E Not enough information is given.
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33 If you are provided with 50 molecules of H2 and 50 molecules of O2, which is the limiting reagent for the reaction:2 H2 (g) + O2 (g) --> 2 H2O (l)
A 50 molecules H2
B 50 molecules O2
C 50 molecules H2O
D There is no limiting reagent.
E Not enough information is given.
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34 If you are provided with 60 molecules of H2 and 20 molecules of N2, which is the limiting reagent for the reaction:3 H2 (g) + N2 (g) --> 2 NH3 (g)
A 60 molecules H2
B 20 molecules N2
C 50 molecules NH3
D There is no limiting reagent.
E Not enough information is given.
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35 If you are provided with 15 molecules of H2 and 10 molecules of N2, which is the limiting reagent for the reaction:3 H2 (g) + N2 (g) --> 2 NH3 (g)
A 15 molecules H2
B 10 molecules N2
C 20 molecules NH3
D There is no limiting reagent.
E Not enough information is given.
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36 Which of the following is NOT a true statement?
A The amount of product is determined by the limiting reagent.
B A balanced equation is necessary to determine which reagent is limiting.
C Some of the excess reagent is left over after the reaction is complete.
D The reactant that has the smallest given mass is the limiting reagent.
E Adding more of the limiting reagent will cause more product to be made.
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2Al(s) + 3Cl2(g) à 2AlCl327 g 71 g ? g
Sample Problem #12Find the theoretical yield of AlCl3, if 27g Al and 71g Cl2 react.
Sample Problem #12 - Solution
1. See how much AlCl3 can be made from each reactant2. Smaller amount identifies the LR
Limiting Reagent & Theoretical Yield
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2Al(s) + 3Cl2(g) à 2AlCl327 g 71 g ? g
Limiting Reagent & Theoretical Yield
Sample Problem #12 - Solution (con't)27g Al will yield _______ g AlCl3
71g Cl2 will yield _______ g AlCl3
Therefore, the LR is _____________and the theoretical yield is ________ grams of AlCl3
27g Al will yield 134 g AlCl3
71g Cl2 will yield 89 g AlCl3
LR = 71 g Cl2TY = 89 g AlCl3
Answer
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2Al(s) + 3Cl2(g) à 2AlCl3100 g 100 g ? g
Sample Problem #13Find the theoretical yield of AlCl3, if 100g of each Al and Cl2 react.
Sample Problem #13 - Solution
1. See how much AlCl3 can be made from each reactant2. Smaller amount identifies the LR
Limiting Reagent & Theoretical Yield
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2Al(s) + 3Cl2(g) à 2AlCl3100 g 100 g ? g
Limiting Reagent & Theoretical Yield
Sample Problem #13 - Solution (con't)100g Al will yield ______ g AlCl3
100g Cl2 will yield _______ g AlCl3
Therefore, the LR is ____________and the theoretical yield of AlCl3 is ____________ grams.
100g Al will yield 494 g AlCl3
100g Cl2 will yield 125 g AlCl3
LR = 100 g Cl2TY = 125 g AlCl3
Answer
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You may sometimes be asked to calculate how much of the excess reagent is left over, after all of the limiting reagent has been consumed.
The general strategy is rather simple, once you have identified the excess reagent.
How to calculate excess reagent
It does not matter if the amounts are in moles or mass (grams), as long as you are consistent.
Excess Reagent
1) Write down how much of it you have at the start of the reaction.
2) Use stoichiometry (i.e. a balanced equation and the LR) to calculate how much of the excess reagent gets used up.
3) Subtract the answer from Step 2 from Step 1.
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BIG IDEA #1:
The limiting reagent determines the theoretical yield.
BIG IDEA #2:
The limiting reagent determines how much of the excess reagent gets used up.
Limiting Reagent, Excess Reagent & Theoretical Yield
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Limiting Reagent & Excess Reagent
Consider this reaction which we have seen before:
2 H2 (l) + O2 (g) --> 2 H2O (l)
Sample Problem #14If we start with 10 molecules of H2 and 7 molecules of O2, how much of the excess reagent will be left over after the reaction is complete?
O2 H2
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Limiting Reagent & Excess Reagent
Sample Problem #14 - SolutionWe had already determined (in Sample Problem #10) that hydrogen is the limiting reagent.
Therefore, oxygen is the excess reagent.and we must now complete the following:
O2 we start with: ________________O2 we use (based on LR): ________________O2 we have left over: __________________
2 H2 (l) + O2 (g) --> 2 H2O (l)O2 H2
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Limiting Reagent & Excess Reagent
Sample Problem #14 - Solution (con't) O2 we start with: 7 molecules O2
O2 we use(based on LR*): 5 molecules O2
O2 we have left over: 2 molecules O2
2 H2 (l) + O2 (g) --> 2 H2O (l)
2 molecules H2 x 1 molecule O2
10 molecules H2 = 5 molecules O2 get used
1Always start with the LR here!
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Fe is the LR.O2 is the ER.You start with 12 molecules of O2 use up 9 of them, so 3 molecules are left over.
Limiting Reagent & Excess Reagent
Try this one on your own:
4 Fe (s) + 3 O2 (g) ® 2 Fe2O3 (s)
Sample Problem #15If we start with 12 molecules of Fe and 12 molecules of O2, how much of the excess reagent will be left over after the reaction is complete?
Answer
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37 If you start with 16 mol H2 and 18 mol O2, how much of the excess reagent is left over?2 H2 + O2 ® 2 H2O
A 2 mol H2
B 7 mol H2
C 2 mol O2
D 10 mol O2
E Nothing is left over.
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38 If you start with 9 mol H2 and 5 mol N2, how much of the excess reagent is left over?
3 H2 + N2 ® 2 NH3
A 2 mol N2
B 4 mol N2
C 4 mol H2
D 6 mol H2
E Nothing is left over.
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39 If you start with 16 mol Fe and 12 mol O2, how much of the excess reagent is left over?
4 Fe + 3 O2 ® 2 Fe2O3
A 4 mol FeB 12 mol FeC 4 mol O2
D 9 mol O2
E Nothing is left over.
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____H2 + ____O2 ---> ____H2O
A. Calculate the theoretical yield (in grams) of water if 6 g H2 and 160 g O2 are combined in a reaction vessel.
B. How many grams of the excess reagent will be remaining in the vessel?
Limiting Reagent & Theoretical YieldPractice Problem 1
6 g H2 is the LR.160 g O2 is the ER.You start with (5 mol) 160 g of O2 and use up (1.5 mol) 48 g, so (3.5 mol) 112 g are left over.
Answer
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200 g AgNO3 is the LR.
7.10 x 1023 atoms Ag can be made.
According to the balanced chemical equation, how many atoms of silver will be produced from combining 100 g of copper with 200 g of silver nitrate?
Cu(s) + 2 AgNO3(aq) à Cu(NO3)2(aq) + 2 Ag(s)
Limiting Reagent & Theoretical YieldPractice Problem 2 *
Answer
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At STP, what volume of “laughing gas” (dinitrogen monoxide) will be produced from 50 g of nitrogen gas and 75 g of oxygen gas? First, write a balanced equation.
Limiting Reagent & Theoretical YieldPractice Problem 3 *
50 g N2 is the LR.
40 L N2O can be made.
Answer
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___N2(g) + ___O2(g) à ___N2O5(g)
A. If you begin with 400 g of N2 and 800 g of O2, how many liters of N2O5 could be produced at STP?
B. Find the mass (in grams) of excess reagent left over at the conclusion of the reaction.
Limiting Reagent & Theoretical YieldPractice Problem 4
800 g O2 is the LR.400 g N2 is the ER.You start with (14.3 mol) 400 g of N2 and use up (10 mol) 280 g, so (4.3 mol) 120 g are left over.
Answer
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Carbon monoxide can be combined with hydrogen to produce methanol, CH3OH. Methanol is used as an industrial solvent, as a reactant in some synthesis reactions, and as a clean-burning fuel for some racing cars. If you had 152.5 kg of carbon monoxide and 24.5 kg of hydrogen gas, how many kilograms of methanol could be produced? (First, write a balanced equation.)
Limiting Reagent & Theoretical YieldPractice Problem 5
152.5 kg CO is the LR.
5446 mol = 174 kg CH3OHcan be made.
Answer
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Credit to Tom GreenboweChemical Education Group at Iowa State University
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Theoretical Yield & Actual Yield
We have recently seen that whenever you are given TWO amounts of reactants, you must first determine which is the limiting reagent.
BIG IDEA: The limiting reagent determines the
theoretical yield.
However, when experiments are carried out in the laboratory, the theoretical yield is not always easily obtained. The amount of product that is made when performing an experiment is called "Actual yield."
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Theoretical Yield vs. Actual Yield
Theoretical yield (TY) Actual yield (AY)
the maximum amount of product that can be made,
based on the stoichiometry (i.e. balanced equation) and
limiting reagent
the amount of product one actually produces and
measures in the laboratory; usually less than the TY
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· Percent yield is the ratio comparing the amount actually obtained (actual yield) to the maximum amount that was possible (theoretical yield).
· The efficiency of a reaction can be expressed using this ratio.
· For example, a percent yield of 85% shows that the reaction conditions are more favorable than with a percent yield of only 55%.
· Chemical manufacturers strive for high percent yields in their processes.
Actual YieldTheoretical Yield Percent Yield = x 100
Percent Yield
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Sample Problem #16Use the balanced equation to determine the percent yield if 5 g Mg reacts to actually form 8.1g of MgO
2 Mg (s) + O2 (g) à 2 MgO (s)
Sample Problem #16 - Solution
1) Since only one amount is given, then 5 g Mg must be the limiting reagent; oxygen is the excess reagent since there is no amount given.
2) Use 5 g to calculate the TY of MgO.
Percent Yield
5 g Mg24.3 g Mg
x 1 mol Mg
= 8.29 g MgO can be made1 2 mol Mg
x 2 mol MgO
1 mol MgO40.3 g MgO
x
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2 Mg(s) + O2(g) à 2 MgO(s)
Sample Problem #16 - Solution (con't)
3) Use percent yield formula
Actual YieldTheoretical Yield Percent Yield = x 100
8.1 g MgO8.3 g MgO Percent Yield = x 100
Percent Yield = 98%
Percent Yield
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40 What is the percent yield if the theoretical yield is 57.3 g and the actual yield is 50.9 g?
A 0.8883%
B 1.1257%C 88.83%D 112.57%
E 0.117%
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41 What is the actual yield if the theoretical yield is 16 g and the percent yield is 94.23%?
A 16.98 g
B 15.08 gC 100 gD 28%
E 1507.68
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Calcium hydroxide, Ca(OH)2, is also known as “slaked lime” and it is produced when water reacts with “quick lime,” CaO. If you start with 2.4 kg of quick lime, add excess water, and produce 2.06 kg of slaked lime, what is the percent yield of the reaction?
Stoichiometry Practice ProblemPractice Problem 6
· Is this a limiting reagent problem?· Is the 2.06 kg a theoretical yield or actual yield?· What quantity must you solve for?· Did you write a balanced equation?
TY = 3.17 kg Ca(OH)2
PY = (2.06/3.17) x 100= 65%
Answer
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TY = 533 g Fe
PY = (464 g Fe/533 g Fe) x 100= 65%
Some underwater welding is done via the thermite reaction, in which rust (Fe2O3) reacts with aluminum to produce iron and aluminum oxide (Al2O3).
In one such reaction, 258 g of aluminum and excess rust produced 464 g of iron. What was the percent yield of the reaction?
Remember to first write a balanced chemical equation.
Stoichiometry Practice ProblemPractice Problem 7
Answer
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TY = 5.6 x 104 L SO2
75% = (AY/5.6 x 104 L) x 100AY = (0.75)(5.6 x 104 L) AY = 4.2 x 104 L SO2
Use the balanced equation to find out how many liters of sulfur dioxide are actually produced at STP if 1.5 x 1027 formula units of zinc sulfide are allowed to react with excess oxygen and the percent yield is 75%.
2 ZnS(s) + 3 O2(g) à 2 ZnO(s) + 2 SO2(g)
Stoichiometry Practice ProblemPractice Problem 8
Answer
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Definitions· Limiting reagent - any reactant that is used up firsts in a chemical reaction; it determines the amount of product that can be formed in the reaction
· Excess reagent - a reagent present in a quantity that is more than sufficient to react with a limiting reagent; any reactant that remains after the limiting reagent is used up in a chemical reaction
· Theoretical yield - the maximum amount of product that can be made, based on the limiting reagent and a balanced chemical equation
· Actual yield - the amount of product that forms when a reaction is carried out in the laboratory
· Percent yield - the ratio of the actual yield to the theoretical yield for a chemical reaction expressed as a percent; it is a measure of the efficiency of a reaction
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Outline:· General idea of stoichiometry· mole ratios· mole-mole problems· particle-particle problems· volume-volume· mole-mole-mass problems· mass-mass problems· mixed problems· LR/TY/PY· how to determine LR· using LR to obtain TY· calculate amount of ER left over· solve for PY or AY