chp 4 - chemical qty and aq. rxn part i(1)
DESCRIPTION
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Chemical Reactions
Stoichiometry is the numerical relationship between chemical quantities
in a balanced chemical equation
Stoichiometry allows for predictions to be made on:
• The amount of products that form based upon the amount of reactants present
• How much of the reactants is needed to form a given amount of product
• How much of one reactant is needed to completely react with another reactant
These calculations are important in chemistry as it allows chemists to plan
and conduct chemical reactions to obtain products in the desired quantities
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StoichiometryMoles of Elements from Compounds
Recall, how we used the chemical formula to determine the ratio of an element in a compound
Example: Calculate the mass, in grams, of oxygen in 126.5 g of sulfuric acid (H2SO4)?
Given: 1 mol H2SO4 = 98.078 g H2SO4
Molar Mass (M.M.) of O = 15.999 g/mol
Step 1: Determine the ratio of oxygen to sulfuric acid
• 4 mol O: 1 mol H2SO4
Step 2: Perform conversion
126.5 𝑔 𝐻2𝑆𝑂4 ×1𝑚𝑜𝑙 𝐻2𝑆𝑂4
98.078 𝑔 𝐻2𝑆𝑂4×
4 𝑚𝑜𝑙 𝑂
1 𝑚𝑜𝑙 𝐻2𝑆𝑂4×15.999 𝑔 𝑂
1 𝑚𝑜𝑙 𝑂= 82.54 𝑔 𝑂
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MOLES of
CompoundMOLES of
Element
MASS of
Compound
MASS of
Element
StoichiometryMoles of Reactants & Products
Now, we will use a balanced chemical equation to determine the amount of a particular reactant need or product that will be produced
A balanced chemical equation is like a “recipe” showing how much reactants are needed to form a desired product
Example: How many moles of NH3 is produced from the complete reaction of 2.4 mol of N2?
Write balanced chemical equation
3𝐻2 (𝑔) + 𝑁2(𝑔) → 2𝑁𝐻3(𝑔)
Determine molar ratio of N2 to NH3
• 1 mol N2 : 2 mol NH3
Perform calculation
2.4 𝑚𝑜𝑙 𝑁2 ×2 𝑚𝑜𝑙 𝑁𝐻3
1 𝑚𝑜𝑙 𝑁2= 4.8 𝑚𝑜𝑙 𝑁𝐻3
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MOLES of
Compound AMOLES of
Compound B
Stoichiometry:Mass of Reactants & Products
Since mole-to-mole conversions have been performed, now mass-to-mass conversions can also be performed
Example: How many grams of NH3 is produced from the complete reaction of 64.2 g of N2?
Write balanced chemical equation:
3𝐻2 (𝑔) + 𝑁2(𝑔) → 2𝑁𝐻3(𝑔)
Determine molar ratio of N2 to NH3
• 1 mol N2 : 2 mol NH3
Perform calculation
64.2 𝑔 𝑁2 ×1𝑚𝑜𝑙 𝑁2
28.014 𝑔 𝑁2×2 𝑚𝑜𝑙 𝑁𝐻3
1 𝑚𝑜𝑙 𝑁2×17.031 𝑔 𝑁𝐻3
1 𝑚𝑜𝑙 𝑁𝐻3= 78.1 𝑔 𝑁𝐻3
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MOLES of
Compound AMOLES of
Compound B
MASS of
Compound A
MASS of
Compound A
Limiting Reactant, Theoretical
Yield & Percent Yield
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𝑃𝑒𝑟𝑐𝑒𝑛𝑡 𝑌𝑖𝑒𝑙𝑑 =𝐴𝑐𝑡𝑢𝑎𝑙 𝑌𝑖𝑒𝑙𝑑
𝑇ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑌𝑖𝑒𝑙𝑑× 100%
Limiting reactant
• The reactant
that makes the
least amount of
product in a
reaction
Theoretical yield
• The amount of
product that can
be made in a
chemical
reaction based
upon the
amount of the
limiting
reactant/reagent
Actual yield
• The amount of
product that is
actually
produced by the
chemical
reaction
Percent yield
• The percentage
of the
theoretical yield
that was actually
attained can be
calculated using
the formula
Limiting Reactant, Theoretical Yield &
Percent YieldExample
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Limiting reactant
Theoretical yield
Actual yield
Percent yield
• If I have 4 crusts and 10 c.
cheese, which is the
limiting reactant?
• How many pizzas can we
expect to make?
• We only make three
pizzas
•3 𝑝𝑖𝑧𝑧𝑎𝑠
4 𝑝𝑖𝑧𝑧𝑎𝑠× 100 = 75%
1 crust + 5 oz. tomato sauce + 2 c. cheese = 1 pizza
Limiting Reactant, Theoretical Yield &
Percent YieldWord Problem Starting with Moles
Example: What is the limiting reactant and theoretical yield of NH3 in moles if you begin with 2.6 mol of hydrogen and 1.4 mol of nitrogen?
1. Write balanced chemical equation:
3𝐻2 (𝑔) + 𝑁2(𝑔) → 2𝑁𝐻3(𝑔)
2. Determine molar ratio of N2 to NH3
• 3 mol H2 : 1 mol N2 : 2 mol NH3
3. Determine limiting reactant and theoretical yield
𝐻2: 2.6 𝑚𝑜𝑙 𝐻2 ×2 𝑚𝑜𝑙 𝑁𝐻3
3 𝑚𝑜𝑙 𝐻2= 1.7 𝑚𝑜𝑙 𝑁𝐻3
𝑁2: 1.4 𝑚𝑜𝑙 𝑁2 ×2 𝑚𝑜𝑙 𝑁𝐻3
1 𝑚𝑜𝑙 𝑁2= 2. 8 𝑚𝑜𝑙 𝑁𝐻3
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MOLES of
Compound AMOLES of
Compound B
Limiting Reactant, Theoretical
Yield & Percent YieldWord Problem Starting with Mass
Example: What is the limiting reactant and theoretical yield of NH3 in moles if you begin with 46.2 g of hydrogen and 76.9 g of nitrogen?
1. Write balanced chemical equation:
3𝐻2 (𝑔) + 𝑁2(𝑔) → 2𝑁𝐻3(𝑔)
2. Determine molar ratio of N2 to NH3
• 3 mol H2 : 1 mol N2 : 2 mol NH3
3. Determine limiting reactant and theoretical yield
𝐻2: 46.2 𝑔 𝐻2 ×1 𝑚𝑜𝑙 𝐻2
2.02 𝑔 𝐻2×2 𝑚𝑜𝑙 𝑁𝐻3
3 𝑚𝑜𝑙 𝐻2= 15.2 mol 𝑁𝐻3 ×
17.03 𝑔 𝑁𝐻3
1 𝑚𝑜𝑙 𝑁𝐻3= 260. 𝑔 𝑁𝐻3
𝑁2: 76.9 𝑔 𝑁2 ×1 𝑚𝑜𝑙 𝑁2
28.014 𝑔 𝑁2×2 𝑚𝑜𝑙 𝑁𝐻3
1 𝑚𝑜𝑙 𝑁2= 5.49 mol 𝑁𝐻3 ×
17.031 𝑔 𝑁𝐻3
1 𝑚𝑜𝑙 𝑁𝐻3= 93.5 𝑔 𝑁𝐻3
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Limiting Reactant
Theoretical Yield
MOLES of
Compound AMOLES of
Compound B
MASS of
Compound A
MASS of
Compound A
Solutions
Most chemical reactions are those
in which reactants are dissolved in
water
Terms:
Solution - a homogeneous mixture
of two or more substances
Solvent - the majority component
of a solution; amount varies
Solute - the minority component
of a solution; amount varies
Aqueous Solution - a solution
where water is the solvent
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Solution Concentration
Quantifying the amount of solute relative to the solvent is to determine
the concentration of the solution
Recall, that the amount of solute and solvent varies
Dilute Solution - contains a small amount of solute relative to the solvent
Concentrated Solution - contains a large amount of solute relative to the
solvent
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Solution ConcentrationMolarity
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Quantify the amount of solute relative to the solvent is to determine
the concentration of the solution
One way to express concentration of a solution is Molarity (M or mol/L)
Molarity is moles of solute per liter of solution
𝑀𝑜𝑙𝑎𝑟𝑖𝑡𝑦 𝑀 =𝑎𝑚𝑜𝑢𝑛𝑡 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 (𝑚𝑜𝑙𝑒)
𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 (𝐿)
Note: Molarity is a ratio of the amount of solute per liter of solution,
not per liter of solvent
Solution ConcentrationCalculating Molarity
Example: Determine the molarity of a solution that has 25.5 g KBr
dissolved to a volume of 1.75 L.
1. Determine formula mass of KBr and convert to moles
25.5 𝑔 𝐾𝐵𝑟 ×1 𝑚𝑜𝑙 𝐾𝐵𝑟
119.00 𝑔 𝐾𝐵𝑟= 0.214 𝑚𝑜𝑙 𝐾𝐵𝑟
2. Calculate the molarity of the solution
𝑀 =0.214 𝑚𝑜𝑙 𝐾𝐵𝑟
1.75 𝐿= 0.122 𝑀 𝐾𝐵𝑟 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
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Preparing a Solution
Can be performed in two
ways:
1. Dissolve a solid solute
with a liquid solvent
2. Mixing a liquid solute
with a liquid solvent
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Preparing a SolutionSolid Solute and Liquid Solvent
Example: Prepare a 2.80 M CaCl2 solution in a 500.0 mL volumetric flask. How many grams of CaCl2 is needed?
1. Molarity is in mol/L so we need our volume in L
500.0 𝑚𝐿 ×1 𝐿
103 𝑚𝐿= 0.5000 𝐿
2. Determine moles of solute (CaCl2) to add to 0.5000 L of solution to make 2.80 M
2.80 𝑚𝑜𝑙 𝐶𝑎𝐶𝑙21 𝐿
× 0.5000 𝐿 = 1.40 𝑚𝑜𝑙 𝐶𝑎𝐶𝑙2
3. Determine amount of solute in grams
1.40 𝑚𝑜𝑙 𝐶𝑎𝐶𝑙2 ×110.984 𝑔
1 𝑚𝑜𝑙 𝐶𝑎𝐶𝑙2= 155 𝑔 𝐶𝑎𝐶𝑙2
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Preparing a Solution
Can be performed in two
ways:
1. Dissolve a solid solute
with a liquid solvent
2. Mixing a liquid solute
with a liquid solvent
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Preparing a SolutionLiquid Solute and Liquid Solvent
Commonly, the solutions that are stored in a lab or purchased are highly concentrated
These solutions are referred to as stock solutions
Will need to make solutions of lower concentrations from stock solutions
To do this, you will need to add more solvent
Note: amount of solute, in moles, does not change, just the volume; thus, moles of solute in original solution is equal to the moles of solute in diluted (new) solution the only thing that changes is the concentration
Equation used to dilute a stock solution
𝑀1𝑉1 = 𝑀2𝑉2or
𝐶1𝑉1 = 𝐶2𝑉2
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M1 or C1 is the concentration of the original solution
V1 is the volume of M1 or C1 (solute) delivered
M2 or C2 is the concentration of the new solution you are trying to make
V2 is the volume of the new concentration (M2 or C2)
Preparing a solution
Example: A 0.50 M NaOH solution is prepared from 50.0 mL of 3.0 M
What is the volume (mL) of the diluted solution?
1. Input the given values into the dilution equation and solve for V2
𝑀1𝑉1 = 𝑀2𝑉2
3.0 𝑀 × 50.0 𝑚𝐿 = 0.50 𝑀 × 𝑉23.0 𝑀 × 50.0 𝑚𝐿
0.50 𝑀= 𝑉2 = 75 𝑚𝐿
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M1 or C1 is the concentration of the original solution
V1 is the volume of M1 or C1 (solute) delivered
M2 or C2 is the concentration of the new solution you are trying to make
V2 is the volume of the new concentration (M2 or C2)
Preparing a solution
Example: Ana prepared 250. mL of a diluted NaOH solution using 8 mL of
a 1.8 M NaOH solution. What was the concentration of her new solution?
1. Input the given values into the dilution equation and solve for V2
𝑀1𝑉1 = 𝑀2𝑉2
1.8 𝑀 × 8 𝑚𝐿 = 𝑀2 × 250 𝑚𝐿
1.8 𝑀 × 8 𝑚𝐿
250 𝑚𝐿= 𝑀2 = 0.06 𝑀
Note: The units used on one side of the dilution equation carry-over to
the other side of the equation
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Solution
stoichiometry
When doing calculations, molarity can be used to convert between the amount of reactants and/or products in a chemical reaction
Example: What volume (in L) of 0.150 M KCl solution is required to completely react with 0.150 L of a 0.175 M Pb(NO3)2 solution according to the following balanced equation?
2𝐾𝐶𝑙(𝑎𝑞) + 𝑃𝑏(𝑁𝑂3)2(𝑎𝑞)→ 𝑃𝑏𝐶𝑙2(𝑠) + 2𝐾𝑁𝑂3(𝑎𝑞)
1. Determine ration between of moles of Pb(NO3)2 and KCl
1 𝑚𝑜𝑙𝑒 𝑃𝑏(𝑁𝑂3)2∶ 2 𝑚𝑜𝑙𝑒 𝐾𝐶𝑙
2. Use the information given to determine the number of moles in Pb(NO3)2
𝑚𝑜𝑙𝑒𝑠 𝑜𝑓𝑃𝑏(𝑁𝑂3)2=0.175 𝑚𝑜𝑙𝑒
𝐿× 0.150 𝐿 = 0.02625 𝑚𝑜𝑙𝑒𝑠
3. Determine volume of KCl solution needed to get necessary moles
0.02625 𝑚𝑜𝑙𝑒𝑠 𝑃𝑏(𝑁𝑂3)2 ×2 𝑚𝑜𝑙𝑒 𝐾𝐶𝑙
1 𝑚𝑜𝑙𝑒 𝑃𝑏(𝑁𝑂3)2×
1 𝐿
0.150 𝑚𝑜𝑙 𝐾𝐶𝑙= 0.350 𝐿 𝐾𝐶𝑙
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MOLES of A MOLES of BVOLUME
of A
VOLUME
of B