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Stoichiometric Calculations Stoichiometry is the quantitative determination of reactants or products using a balanced chemical equation.
Stoichiometric
Calculations
We can interpret a balanced equation in several ways.
One way is to interpret it as indicating the number of moles of each substance. For instance, this formula 2 H2 + O2 → 2 H2O
can be read as:
2 moles of H2 plus 1 mole of O2 yields 2 moles of H2O The coefficients in the balanced equation give the ratio of moles of reactants and products.
Stoichiometric Calculations with Moles Using this interpretation it's straightforward to answer questions about the relative number of moles of reactants and products.
For instance, use the balanced equation below to determine the maximum number of moles of H2O that could be created from reacting 8 moles of H2.
2 H2 + O2 → 2 H2O
Stoichiometry Calculations with Moles 2 H2 + O2 → 2 H2O
1. Use the equation to set up a ratio of the substances of interest.
2 mol H2O
2 mol H2
2. Set that equal to the ratio of the known to unknown quantities of the same substances.
2 mol H2O
n mol H2O
=
8 mol H2
2 mol H2
3. Solve for the unknown.
n = 8.0 moles
Stoichiometry Calculations with Moles Stoichiometry Calculations with Moles The same approach can tell you how many moles of one reactant is needed to completely react with another reactant.
For instance, use the below formula to determine the number of moles of O2 that will be necessary to completely react with 12 moles of H2.
2 H2 + O2 → 2 H2O
2 H2 + O2 → 2 H2O
1. Use the formula to set up a ratio of the substances of interest.
1 mol O2
2 mol H2
2. Set that equal to the ratio of the known to unknown quantities of the same substances.
n mol O2 1 mol O2
=
12 mol H2 2 mol H2
3. Solve for the unknown.
n = 6 mol
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What is the largest number of moles of Al2O3 that could result from reacting 6 moles of O2?
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How many moles of O2 would be required to create 12 moles of Al2O3? 2
4 Al (s) + 3 O2 (g) → 2 Al2O3 (s)
4 Al (s) + 3 O2 (g) → 2 Al2O3 (s)
How many moles of O2 would be required to completely react with 8 moles of Al? 3
4 Al (s) + 3 O2 (g) → 2 Al2O3 (s)
4
When iron rusts in air, iron (III) oxide is produced. How many moles of oxygen react with 2.4 mol Fe in this reaction? 4 Fe (s) + 3 O2 (g) ­­> 2 Fe2O3 (s)
5
How many moles of aluminum are needed to react completely with 1.2 mol FeO? 6
2 Al (s) + 3 FeO (s) ­­> 3 Fe (s) + Al2O3 (s)
How many moles of calcium metal are produced from the decomposition of 8 mol of calcium chloride? CaCl2 (s) ­­> Ca (s) + Cl2 (g)
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How many moles (total) of calcium metal and chlorine gas are produced from the decomposition of 8 mol of calcium chloride? March 19, 2014
Stoichiometric Calculations
A balanced chemical equation is needed to perform any stoichiometric calculations.
CaCl2 (s) ­­> Ca (s) + Cl2 (g)
N2 + 3H2 ­­­> 2NH3
1 mol N2 3 mol H2 1 mol N2 3 mol H2 2 mol NH3 2 mol NH3 The above ratios can be used to determine the quantity of any reactant and products.
For every 1 mol of N2,
• you would need 3 mol of H2 to completely react with
• you would produce 2 mol of NH3 Stoichiometric Calculations
Sample Problem 1
Stoichiometric Calculations
Solution 1A
N2 + 3H2 ­­­> 2NH3
How many moles of hydrogen gas are needed to produce 17 mol ammonia? (assuming an unlimited amount of nitrogen gas available.)
1 mol N2 3 mol H2 1 mol N2 3 mol H2 2 mol NH3 2 mol NH3 How many moles of hydrogen gas are needed to produce 17 mol ammonia?
Method A
Create a mole ratio that involves the "wanted" quantity in the numerator and the "given" quantity in the denominator.
b mol of "wanted" substance W
a mol of "given" substance G
Consider the quantity that is given (ammonia) and the one that is asked for (hydrogen). The second ratio is the one we shall use because it incorporates both of these substances. a and b are coefficients
aG
bW
given quantity
wanted quantity
Stoichiometric Calculations
Stoichiometric Calculations
Solution 1B
Solution 1A
How many moles of hydrogen gas are needed to produce 17 mol ammonia?
Method A
Always write the given quantity first. Then, multiply it by the mole ratio you created.
b mol W x b x mol G x = mol W
a mol G a
Method B
Use a mole ratio that involves the given quantity and wanted quantity. Set up a proportion and solve by cross­
mutliplying.
3 mol H2 =
2 mol NH3 given mole ratio calculated
17 mol NH3 x
3 mol H
2 2 mol NH3 = 25.5 mol H2
X = 17 mol NH3 x
x mol H2
17 mol NH3 3 mol H2 2 mol NH3 = 25.5 mol H2
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Stoichiometric Calculations
Warm Up
How many moles of lithium nitride can be produced from 5 moles of lithium reacting with an unlimited amount of nitrogen gas.
6 Li (s) + N2 (g) → 2 Li3N (s)
Mole­Mole Practice 1
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?
B. From the amount of nitric acid given in Part A, how many moles of silver nitrate will be produced?
C. From the amount of nitric acid given in Part A, how many moles of water will be produced?
D. From the amount of nitric acid given in Part A, how many moles of nitrogen monoxide will be made? Stoichiometric Calculations
Mole­Mole Practice 2
2 N2H4(l) + N2O4(l) à 3 N2(g) + 4 H2O(g)
Stoichiometry Calculations with Particles
A second interpretation of a balanced equation is in terms of the particles (which could be molecules, atoms, or formula units). A. How many moles of dinitrogen tetrahydride are required to produce 57 moles of nitrogen?
Dividing 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. Then this B. How many moles of dinitrogen tetroxide are required to produce 57 moles of nitrogen?
can be read as:
C. How many moles of water are produced when 57 moles of nitrogen are made?
2 H2 + O2 → 2 H2O
2 molecules of H2 plus 1 molecule of O2 yields 2 molecules of H2O.
Not very different. But, while you can get answers that are fractions of moles...particles must be whole numbers.
8
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)
9
How many N2 molecules would be required to create 4 Li3N formula units? 6 Li (s) + N2 (g) → 2 Li3N (s)
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Stoichiometry Calculations with Gases
10 How many Li atoms would be required to completely react with 3 N2 molecules? A third interpretation applies if there are only gases involved. If all the gases are at the same temperature and pressure, the coefficients give the relative volumes of gas. 6 Li (s) + N2 (g) → 2 Li3N (s)
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
This approach only works when comparing two gases in a formula. It cannot be used to compare a gas to a liquid or solid.
[ * at STP conditions:
1 x 22.4 L + 3 x 22.4 L → 1 x 22.4 L + 2 x 22.4 L
]
Stoichiometry Calculations with Gases
Stoichiometry Calculations with Gases
CS2 (g) + 3 O2 (g)→ CO2 (g) + 2 SO2 (g)
Using this interpretation we can directly answer questions about the relative volumes of gas reactants and products.
1. Use the formula to set up a useful ratio.
2 L of SO2
3 L of O2
For instance, how many liters of SO2 will be produced when 6 liters of O2 are reacted as shown below?
2. Set that equal to the ratio of the known to unknown quantities of the same substances.
CS2 (g) + 3 O2 (g)→ CO2 (g) + 2 SO2 (g)
2 L of SO2
V of SO2
=
3 L of O2
6 L of O2
3. Solve for the unknown.
V of SO2 = 6 (2/3) L = 4 L
The coefficients in the balanced equation give the ratio of moles of reactants and products.
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What volume of methane is needed to completely react with 500 L of O2 at STP?
__ CH4 + ___ O2 −−> ___ CO2 + ___ H2O 12
What volume of methane is needed to completely react with 500 L of O2 at STP?
__ CH4 + ___ O2 −−> ___ CO2 + ___ H2O 5
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14 How many liters of H2O (g) will be The equation below shows the created from reacting 8.0 L of H2 (g) with a sufficient amount of O2 (g)?
13 decomposition of lead nitrate. How many liters of oxygen are produced when 12 L NO2 are formed? 2 H2 (g) + O2 (g) → 2 H2O (g)
2Pb(NO3)2(s) → 2PbO(s) + 4NO2(g) + O2(g) V of H2O (g)
=
8 L of H2 (g)
2 L of H2O (g)
2 L of H2 (g)
V of H2O (g) = 8L(2/2) = 8 L
Stoichiometry Calculations in General
How many liters of NO (g) will be 2
15 created from reacting 36 L of O
2 (g) with a sufficient amount of NH3 (g)?
If there is a mixture of gases and non­gases in a formula, we can't directly determine the volume of gas generated. 4 NH3 (g) + 7 O2 (g) → 4 NO2 (g) + 6 H2O (g)
4 L of NO2 (g)
V of NO2 (g)
=
7 L of O2 (g)
36 L of O2 (g)
However, we can determine that by first calculating the number of moles of gas generated and then converting that to a Volume. V of NO2 (g) = 36 L (4/7) = 21 L
At STP, we know that each mole of gas will fill a volume of 22.4 L. In these general cases, we may need to convert the given masses, moles and/or volumes back to either moles or masses to solve the problem. The coefficients in the balanced equation give the ratio of moles of reactants and products.
Stoichiometry Stoichiometry Coefficients in a balanced equation can represent either moles or particles (atoms, molecules, or formula units). As we have seen, we can interpret a balanced equation in several ways. By multiplying by MW, the formula coefficients relate masses. 2H2 + O2 −−> 2H2O
2H2 + O2 −−> 2H2O
2 molecules H2 + 1 molecule O2 −−> 2 molecules H2O
2 molecules H2 + 1 molecule O2 −−> 2 molecules H2O
2 mol H2 + 1 mol O2 −−> 2 mol H2O
2 mol H2 + 1 mol O2 −−> 2 mol H2O
4.0 amu H2 + 32.0 amu O2 −−> 36.0 amu H2O
4.0 amu H2 + 32.0 amu O2 −−> 36.0 amu H2O
4.0 g H2 + 32.0 g O2 −−> 36.0 g H2O
4.0 g H2 + 32.0 g O2 −−> 36.0 g H2O
For formulas with only gases, the coefficients also represent volumes.
The coefficients in the balanced equation give the ratio of moles of reactants and products.
2 L H2 (g)
+ 1 L O2 (g) → 2 L H2O (g)
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Stoichiometry Which of the following statements is 3H2 + N2 à 2NH3
16 true for this equation showing the complete combustion of pentane? 3H2
N2
2NH3
6 atoms of hydrogen
2 atoms of nitrogen
2 atoms of nitrogen and 6 atoms of hydrogen
3 molecules of hydrogen
1 molecule of nitrogen
2 molecule of ammonia
x 10= 30 molecule of H2
x10=10 molecules of N2
x10= 20 molecule of NH3
3 x 6.02.1023 molecule of H2
1 x 6.02.1023 molecule of N2
2 x 6.02.1023 molecule of NH3
3 mole of H2
1 mole of N2
2 mole of NH3
3 x 2 = 6g of hydrogen
1 x 28g = 28g of N2
2 x (14+3)= 34g of ammonia
3 x 22.4 L of hydrogen
1 x 22.4L of nitrogen
2 x 22.4L of ammonia
17 Which of these is an incorrect interpretation of this balanced equation? C5H12 (l) + 8 O2 (g) ­­> 5 CO2 (g) + 6 H2O (g)
18
In any chemical reaction the quantities that are preserved are _____. 2 S (s) + 3 O2 (g) ­­> 2 SO3 (g)
Warm Up
How many moles of lithium nitride can be produced from 5 moles of lithium reacting with an unlimited amount of nitrogen gas.
6 Li (s) + N2 (g) → 2 Li3N (s)
Stoichiometric Calculations with Mass
So far, we have seen stoichiometry problems that can be solved in one step, using only coefficients. These are: mole­mole problems, •
particle­particle problems and •
volume­volume problems. •
When mass is involved, we must now take into consideration the fact that all particles have a different molar mass. Coefficients do not represent mass in grams.
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Mass­Mass Calculations
Mass­Mass Calculations
Starting with the mass of Substance A you can use the ratio of the coefficients of A and B to calculate the mass of Substance B formed (if it’s a product) or used (if it’s a reactant).
aA bB
Given
Sample Problem 1
Calculate the mass of ammonia, NH3, produced by the reaction of 5.4 g hydrogen gas with an excess of nitrogen.
N2 + 3H2 ­­­> 2NH3
Find
Grams of substance A
Grams of substance B
Using Method A, we proceed using these steps:
a)
b)
c)
g H2 ­­> mol H2 ­­> mol NH3 ­­> g NH3
Use molar
mass of A
Use molar
mass of B
Use coefficients of A and B from
balanced equation
Moles of substance A
a) convert 5.4 g H2 to moles H2
b) convert moles H2 to moles NH3
c) convert moles NH3 to grams NH3
Moles of substance B
Mass­mass Calculations
Mass­Mass Calculations
Sample Problem 2
Sample Problem 1
mol H2 −−> mol NH3 −−> g NH3
gH2 −−>
How many grams of water can be produced from the combustion of 1.00 g of glucose?
C6H12O6 + 6 O2 ­­> 6 CO2 + 6 H2O
Method A
1 mol H2
5.40 g H2 X
2.0 g H2
2 mol NH3
X
3 mol H2
X
17.0 g NH3
1 mol NH3
= 31 g NH3
c) convert moles NH3 to grams NH3
a) convert 5.4 g H2 to moles H2
b) convert moles H2 to moles NH3
Step (a): convert mass of glucose to moles 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
use the molar mass of water
•
Mass­mass Calculations
Sample Problem 2
How many grams of water can be produced from the combustion of 1.00 g of glucose?
C6H12O6 + 6 O2 ­­> 6 CO2 + 6 H2O
no direct calculation
1.00 g glucose
X
(c)
18.0 g water
X
180.0 g glucose
5.56 x 10­3 mol glucose
X
6 mol water
Sample Problem 2
How many grams of water can be produced from the combustion of 1.00 g of glucose?
0.600 g water
(a)
1mol glucose
Mass­mass Calculations
C6H12O6 + 6 O2 ­­> 6 CO2 + 6 H2O
There is an alternative way to solve this problem that requires interpreting the balanced equation in another way.
1 mol water
3.33 x 10­2 mol water
1 mol glucose
(b) for every 1 mol glucose
you get 6 mols of water
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Mass­mass Calculations
Mass­mass Calculations
Sample Problem 2
Sample Problem 2
How many grams of water can be produced from the combustion of 1.00 g of glucose?
How many grams of water can be produced from the combustion of 1.00 g of glucose?
Method B requires first using the balanced equation to write down all the correct mass proportions of substances.
Method B 6 O2
C6H12O6
6 CO2
6 H2O
1 mol C6H12O6
6 mol O2 6 mol CO2 6 mol H2O 180 g C6H12O6
192 g O2 264 g CO2 108 g H2O C6H12O6 + 6 O2 ­­> 6 CO2 + 6 H2O
C6H12O6
1 mol C6H12O6
6 O2
6 mol O2 6 H2O
6 CO2
6 mol CO2 6 mol H2O 6 x 32g = 6 x 44g =
6 x 18g =
1 x 180g =
180 g C6H12O6 192 g O2 264 g CO2 108 g H2O Mass­mass Calculations
How many grams of water can be produced from the combustion of 1.00 g of glucose?
Method B 108 g H2O
=
ratio formed from balanced equation
x g H2O
1 g C6H12O6
Stoichiometry Calculations with Masses
We now see that a fourth interpretation of a balanced equation is in terms of mass. Sample Problem 2
180 g C6H12O6
We choose to create a ratio with the highlighted quantities because these are the two substances that are part of the problem.
x = 0.60 g H2O
ratio using given quantity and unknown 2 H2 + O2 → 2 H2O
A mole of H2 has a mass of 2 x 1.0 g = 2 g; so 2 moles has a mass of 4 g
A mole of O2 has a mass of 2 x 16 g = 32 g;
so 1 mole has a mass of 32 g
A mole of H2O has a mass of 1 x 2 g + 1 x 16g = 18g;
so 2 moles has a mass of 36 g
So our equation can be read as:
4 g of H2 plus 32 g of O2 yields 36 g of H2O Stoichiometry Calculations with Masses
Stoichiometric Calculations
Using this interpretation we can answer questions about the relative masses of reactants and products.
How many grams of Na can be produced from the decomposition of 40 g of NaN3?
For instance, use the equation below formula to determine the mass of Na (s) that results from the decomposition of 40 g of NaN3 (s).
2 NaN3 (s) → 2 Na (s) + 3 N2 (g)
Sample Problem 3A
2 NaN3 (s) → 2 Na (s) + 3 N2 (g)
Method A
a) 40 grams of NaN3 to moles NaN3 b) moles NaN3 to moles Na
c) moles Na to grams Na
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Mass­mass Calculations
Mass­mass Calculations
Sample Problem 3B
Sample Problem 3B
How many grams of Na can be produced from the decomposition of 40 g of NaN3?
2 NaN3 (s) → 2 Na (s) + 3 N2 (g)
130 g of NaN3 → 46 g of Na + 84 g of N2
4. Use the formula to set up a useful ratio.
46 g of Na
130 g of NaN3
1. First, determine the masses of the products and reactants. NaN3 = 1 x 23 g + 3 x 14 g = 65 g; so 2 moles = 130 g
Na = 1 x 23 g = 23 g; so 2 moles = 46 g
N2 = 2 x 14 g = 28 g; so 3 moles = 84 g
5. Set that equal to the ratio of the known to unknown quantities of the same substances.
m of Na
2. Rewrite the formula in terms of mass.
130 g of NaN3 → 46 g of Na + 84 g of N2
40 g of NaN3
created from reacting 36 g of Al with a sufficient amount of O2?
46 g of Na
130 g of NaN3
6. Solve for the unknown.
3. Double check to see that mass of products = mass of reactants 130 g = 46 g + 84 g
19 How many grams of Al2O3 will be =
m of Na = 40 g x (46g/130g) = 14 g
How many grams of Mg must react in 20 order to to create 84 g of MgO? 2 Mg (s) + O2 (g) → 2 MgO (s)
4 Al (s) + 3 O2 (g) → 2 Al2O3 (s)
1. Mg = 1 x 24.3 g = 24.3 g; so 2 moles = 48.6 g
O2 = 2 x 16 g = 32 g; so 1 mole = 32 g
MgO = 1 x 24.3 g + 1 x 16g = 40.3 g; so 2 moles = 80.6 g
1. Al = 1 x 27 g = 27 g; so 4 moles = 108 g
O2 = 2 x 16 g = 32 g; so 3 moles = 96 g
Al2O3 = 2 x 27 g + 3 x 16g = 102 g; so 2 moles = 204 g
2. 48.6 g of Mg + 32 g of O2 → 80.6 g of MgO 2. 108 g of Al + 96 g of O2 → 204 g of Al2O3 3. 48.6 g + 32 g = 80.6 g; it checks
3. 108 g + 96 g = 204 g; it checks
4 & 5. 4 & 5. m of Al2O3 =
36 g of Al
204 g of Al2O3 108 g of Al
m of Mg
=
84 g of MgO
48.6 g of Mg 80.6 g of MgO
6. m of Mg = 84(48.6/80.6) = 51 g
6. m of Al2O3 = 36(204/108) = 68 g
21 What mass of CaO would be required to completely react with 42 g of H2O? CaO (s) + H2O (l) → Ca(OH)2 (s)
1. CaO = 1 x 40 g + 1 x 16 g = 56 g
H2O = 2 x 1.0 g +1 x 16 g = 18 g
Ca(OH)2 = 1 x 40 g + 2 x 16 g + 2 x 1.0 g = 74 g
Mass­mass Calculations
Practice Problem 4
__ CaCl2 (aq) + __ AgNO3 (aq) à __ Ca(NO3)2 (aq) + __ AgCl (s) Balance the equation above. How many grams of silver chloride are produced when 45 g of calcium chloride react with excess silver nitrate?
2. 56 g of CaO + 18 g of H2O → 74 g of Ca(OH)2 3. 56 g + 18 g = 74 g; it checks
4 & 5. m of CaO
=
42 g of H2O
56 g of CaO
18 g of H2O
6. m of CaO = 42(56/18) = 131 g
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Mass­mass Calculations
March 19, 2014
Mixed Stoichiometry Calculations
Practice Problem 5
Fe2O3 (s) + C (s) ­­> Fe (s) + CO2 (g)
Balance the equation above. How many grams of iron can be recovered from 500 kg of the oxide ore, Fe2O3?
Thus far, our problem­solving 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?
Mixed Stoichiometry Calculations
Stoichiometric Calculations
Every type of stoichiometry calculation may be solved by following this map. Mixed Problem 1A
(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.
How many grams of ammonia can be produced by using 10 mol of nitrogen gas and an unlimited or excess amount of hydrogen gas?
N2 + 3H2 ­­­> 2NH3
Method A
(3)
(1)
represenative particles of G X
1 mol G
x
6.02 x10­23
6.02 x 1023
representative
= particles of W
1 mol W
Since the "Given" quantity is alread in moles, we can skip to step (2).
(2)
mass 1 mol G
of G X
mass G
b mol W
mol G X
mol W
mass W
x
1 mol W
a mol G
1 mol G
volume of G X
at STP
22.4 L G
x
22.4 L W
1 mol W
mass
= of W
Overview:
mol N2 ­­­> mol NH3 ­­­> grams NH3
Volume of = W at STP
Stoichiometric Calculations
Stoichiometric Calculations
Mixed Problem 1A
Mixed Problem 1B
How many grams of ammonia can be produced by using 10 mol of nitrogen gas and an unlimited or excess amount of hydrogen gas?
N2 + 3H2 ­­­> 2NH3
Method A
We can approach this same problem using another method.
Method B requires first writing down all the relative proportions of substances using the balanced equation.
N2
10 mol N2 x
How many grams of ammonia can be produced by using 10 mol of nitrogen gas and an unlimited or excess amount of hydrogen gas?
2 mol NH3 17 g NH3
x
1 mol N2 1 mol NH3
= 340 g NH3
3H2
2NH3
1 mol N2
3 mol H2 2 mol NH3
28 g N2
6 g H2
34 g NH3
67.2 L H2 44.8 L NH3 22.4 L N2
Any two of these quantities may be used to create a ratio.
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Stoichiometric Calculations
Mixed Problem 1B
How many grams of ammonia can be produced by using 10 mol of nitrogen gas and an unlimited or excess amount of hydrogen gas?
Method B We choose the two highlighted quantities to create a ratio because they are the quantities that are involved in the problem.
N2
3H2
2NH3
1 mol N2
3 mol H2 2 mol NH3
28 g N2
6 g H2
34 g NH3
67.2 L H2 44.8 L NH3 22.4 L N2
Stoichiometric Calculations
Mixed Problem 2
The 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?
March 19, 2014
Stoichiometric Calculations
Mixed Problem 1B
How many grams of ammonia can be produced by using 10 mol of nitrogen gas and an unlimited or excess amount of hydrogen gas?
Method B
34 g NH3 x g NH3
=
1 mol N2 x = 340 g NH3
10 mol N2 ratio formed from balanced equation
ratio using given quantity and unknown Stoichiometric Calculations
Mixed Problem 2
2 NaN3 (s) → 2 Na (s) + 3 N2 (g)
1. Convert 36 L of N2 to moles: 36 L (1 mole / 22.4 L) = 1.6 moles
2. Use the formula to determine the needed moles of NaN3.
2 mol NaN3
n mol NaN3
=
3 mol N2
1.6 mol N2
3. Solve for the unknown. In this case, let's convert both the quantities of NaN3 that we're solving for and the volume of N2 we're being given to moles. Then we can directly use the above equation. This approach will always yield a result.
n = 1.6 mol (2/3) = 1.1 mol NaN3
4. Determine the mass of 1 mol NaN3 (M) and multiply by n.
M for NaN3 = 1 x 23 g + 3 x 14 g = 65 g/mol
m = nM = (1.1 mol)(65 g/mol) = 72 g
How many liters of O2 (g) at STP are 22 required to create 90 g of Al2O3 (s)?
Stoichiometric Calculations
4 Al (s) + 3O2 (g) → 2 Al2O3 (s)
Mixed Problem 3
2 Sb + 3 Cl2 à 2 SbCl3
M for Al2O3 = 2 x 27 g + 3 x 16 g = 102 g/mol
How many grams of Cl2 are needed to react with 1 mol of Sb?
n = m/M = (90 g)(102 g/mol) = 0.88 mol
3 mol O2
n mol O2
=
0.88 mol Al2O3
2 mol Al2O3
n mol O2 = m/M = (0.88 mol)(3/2) = 1.32 mol O2
V = nVm = (1.32 mol)(22.4 mol/L) = 29.57 L O2
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Stoichiometric Calculations
Mixed Problem 4
3MnO2(s) + 4 Al(s) à 2 Al2O3(s) + 3Mn(s)
A. How many manganese atoms are produced if 54.8 moles of Mn3O4 react with excess aluminum.
March 19, 2014
Stoichiometric Calculations
Mixed Problem 4
3MnO2(s) + 4 Al(s) à 2 Al2O3(s) + 3Mn(s)
C. How many moles of manganese oxide will react with 5.33 x 1025 atoms of aluminum?
B. How many moles of aluminum oxide are made if 3580 g of manganese oxide are consumed?
D. If 4.37 moles of aluminum are consumed, how many molecules of aluminum oxide are produced?
Stoichiometric Calculations
Mixed Problem 5
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:
Stoichiometric Calculations
Mixed Problem 5
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?
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?
D. Find the mass of tristearin required to produce 55.56 moles of water. B. At STP, what volume of carbon dioxide is produced in Part A?
How Many Cookies Can I Make?
Limiting Reagent & Excess Reagent
Percent Yield & Theoretical Yield
• 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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How Many Cookies Can I Make?
• In this example the sugar would be the limiting reactant, because it will limit the amount of cookies you can make.
March 19, 2014
Limiting Reactants
• 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.
23 When two substances react to form products, the reactant which is used up is called the
Limiting Reactants
Limiting reagent problems are different from those done previously in that two quantities of the reactants 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. Two different methods are presented here.
Limiting Reactants
Consider the reaction between hydrogen and oxygen to yield water. 2 H2 (l) + O2 (g) ­­> 2 H2O (l)
Limiting Reactant & Excess Reactant
In this example, ____ is the limiting reagent and ____is the excess reagent.
Starting with 10 molecules of H2 and 7 molecules of O2, which reactant will run out first?
Before reaction
10H2 and 7 O2
After reaction
10 H2O and 2O2
Before reaction
10H2 and 7 O2
After reaction
10 H2O and 2O2
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Limiting Reactants
Method A ­ overview
1. Write a balanced equation.
2. Use the amount of each reactant given to calculate the amount of product that could be formed. 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 creates the smaller amount of product is thus the limiting reagent.
Limiting Reactants
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. March 19, 2014
Limiting Reactants
Limiting reagent problems are different from those done previously in that two quantities are given, instead of just one. There are various methods used to determine which reactant is the limiting one. Consider the previous example.
Method A ­ Example
1. Write a balanced equation.
2 H2 (l) + O2 (g) ­­> 2 H2O (l)
2. Use the amount of each reactant given to calculate the amount of product that could be formed. 10 molecules H2 would theoretically yield 10 molecules water
7 molecules O2 would theoretically yield 14 molecules water
Limiting Reactants
Method B­ overview
1. Write a balanced chemical equation.
In actuality, 10 molecules of water are formed.
The larger of the two amounts is irrelevant and meaningless.
2. Write down both given quantities under the equation (include units).
It is impossible to produce 14 molecules of water, since there is not enough hydrogen gas.
3. Convert all quantities to moles.
4. Whichever reactant creates the smaller amount of product is thus the limiting reagent.
4. Divide each # of moles by the coefficient in front of each reactant. The smallest value is the limiting reagent.
Hydrogen (10 molecules) is the limiting reagent. The theoretical yield, or maximum amount of product that could be made, is 10 molecules of water.
Limiting Reactants
Method B Example: How much ammonia can be produced from 65 g nitrogen and 25 g hydrogen?
1. Write a balanced chemical equation. Write down both given quantities under the equation (include units).
N2 + 3H2 ­­­> 2NH3
65 g
25 g
2. Convert all quantities to moles.
65 g N2 = 2.3 mol N2
25 g H2 = 12.5 mol H2
5. Use the given quantitiy of the LR to determine the theoretical yield.
Limiting Reactants
Method B Example: What mass of ammonia can be produced from 65 g nitrogen and 25 g hydrogen?
N2 + 3H2 ­­­> 2NH3
3. Divide each # of moles by the coefficient in front of each reactant. The smallest value corresponds to the limiting reagent.
2.3 mol N2 / 1 = 2.3
12.5 mol H2 / 3 = 4.2
4. Use the given quantitiy of the LR to determine the theoretical yield. Since N2 is the LR, convert from 2.3 mol N2 to mass of NH3 using the stoichiometric ratio. Answer: 78 g NH3
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24 Which of the following would be the limiting reagent in the reaction: 2H2(g) + O2(g) → 2H2O(g) March 19, 2014
25 Which of the following is NOT a true statement concerning limiting and excess reagents?
Limiting Reagent & Theoretical Yield
Limiting Reagent & Theoretical Yield
Sample Problem 1: Find the theoretical yield of AlCl3, if 27g Al and 71g Cl2 react. Here we will apply Method A.
Sample Problem 2: Find the theoretical yield of AlCl3, if 100g of each Al and Cl2 react. Here, we will apply Method B.
1. See how much AlCl3 can be made from each reactant
2. Smaller amount identifies the LR 2Al(s) + 3Cl2(g) à 2AlCl3
? g
71 g
27 g 1. Find the limiting reagent
2. Use the LR to determine the theoretical yield of AlCl3
2Al(s) + 3Cl2(g) à 2AlCl3
? g
100 g
100 g
27g Al will yield _______ g AlCl3
71g Cl2 will yield _______ g AlCl3
100g Al /27g/mol 100g Cl2/71g/mol
Therefore, the LR is ________ and the theoretical yield of AlCl3 is _________ grams.
=3.70 mol Al = 1.41 mol Cl2
3.70/2 = 1.85 1.41/3 = 0.47
smallest number
Limiting Reagent = Cl2
Limiting Reagent & Theoretical Yield
Limiting Reagent & Theoretical Yield
Practice Problem 1
2Al(s) + 3Cl2(g) à 2AlCl3
100 g
100 g
? g
Now that we know that the Limiting Reagent is Cl2, we can use the amount of Cl2 to determine the theoretical yield of AlCl3.
3 x 71.0g 100g of Cl2 =
2 x 133.35g X g of AlCl3
1.41 mol Cl2 x
2 mol AlCl3
a) Calculate the theoretical yield of water if 6 g H2 and 160 g O2 are combined in a reaction vessel.
OR
b) How many grams of the excess reagent will be remaining in the vessel?
133.5 g AlCl3
x
3 mol Cl2
____H2 + ____O2 ­­­> ____H2O
1 mol AlCl3
=
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March 19, 2014
Limiting Reagent & Theoretical Yield
Limiting Reagent & Theoretical Yield
Warm Up
Practice Problem 2
Solid copper reacts with a solution of silver nitrate. Write a balanced chemical reaction.
What is the coefficient for silver nitrate in the balanced chemical reaction?
Cu(s) + 2 AgNO3(aq) à Cu(NO3)2(aq) + 2 Ag(s)
26 Answer?
Determine the limiting reactant if 100 g of copper reacts with 200 g of silver nitrate.
Cu(s) + 2 AgNO3(aq) à Cu(NO3)2(aq) + 2 Ag(s)
27 A
Cu
B
AgNO3
Limiting Reagent & Theoretical Yield
Limiting Reagent & Theoretical Yield
Practice Problem 2
Practice Problem 3
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?
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.
Cu(s) + 2 AgNO3(aq) à Cu(NO3)2(aq) + 2 Ag(s)
28 Answer?
Limiting Reagent & Theoretical Yield
Practice Problem 4
___N2H4(l) + ___N2O4(l) à ___N2(g) + ___H2O(g).
A. Determine the limiting reagent if you begin with 400 g of N2H4 and 900 g of N2O4.
___N2H4(l) + ___N2O4(l) à ___N2(g) + ___H2O(g).
400 g
900 g
C. What volume of nitrogen is produced at STP, assuming the reaction goes to completion?
B. Find the number of liters of water produced at STP, assuming the reaction goes to completion (i.e. one of the reactants gets completely used up).
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Limiting Reagent & Theoretical Yield
___N2H4(l) + ___N2O4(l) à ___N2(g) + ___H2O(g).
400 g
900 g
D. Find the mass of excess reactant left over at the conclusion of the reaction.
Practice Problem 5
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 clean­burning 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?
Theoretical Yield vs. Actual Yield
Theoretical yield Actual yield the maximum amount of product that can be made, based on the stoichiometry (i.e. balanced equation)
the amount of product one actually produces and measures in the laboratory; usually less than the TY Credit to Tom Greenbowe
Chemical Education Group at Iowa State University
Percent Yield
• The efficiency of a reaction can be expressed as a ratio of the actual yield to the theoretical yield. • For example, a percent yield of 85% shows that the reaction conditions are more favorable than with a percent yield of only 55%. 29 For a given chemical reaction, the theoretical yield is _____ greater than the actual yield. • Percent yield is the ratio comparing the amount actually obtained (actual yield) to the maximum amount that was possible (theoretical yield).
Actual Yield
x 100
Percent Yield = Theoretical Yield
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Stoichiometry Practice Problem
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?
•
•
•
•
March 19, 2014
Stoichiometry Practice Problem
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?
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?
Stoichiometry Practice Problem
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)
Definitions
• Theoretical yield ­ the amount of product that could form during a reaction; it is calculated from a balanced chemical equation and it represents the maximum amount of product that could be formed from a given amount of reactant.
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
Skills Practice 33 Answers
1. a. O2
b. 351 g CO2, 180 g H2O
2. 322 g CaF2, 322 g Al2(CO3)3
• Actual yield ­ the amount of product that forms when a reaction is carried out in the laboratory
3. a. 18.6 g Ca3O2 left over
b. 130 g
• 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
4. 280 g K3P, 142 g Al2S3
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Attachments
Skill Practice 32­34 answers.docx
Stoichiometry Unit Review 2013.doc
Schedule 2­25 2013.doc
ChemQuest 32­34 answers.docx
micromole rockets entire lab 1011.pdf
micromole rockets.notebook
micromole rockets.pdf
Skill Practice 30­34.doc
Lab limiting reagent smore.doc