1 calculations from chemical equations chapter 9 hein and arena version 1.1
TRANSCRIPT
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Calculations FromChemical Equations
Chapter 9
Calculations FromChemical Equations
Chapter 9
Hein and Arena
Version 1.1
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Chapter Outline9.1 A Short Review
9.2 Introduction to Stoichiometry:The Mole-Ratio Method
9.3 Mole-Mole Calculations
9.4 Mole-Mass Calculations
9.5 Mass-Mass Calculations
9.6 Limiting-Reactant and Yield Calculations
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• The molar mass of an element is its atomic mass in grams.
• It contains 6.022 x 1023 atoms (Avogadro’s number) of the element.
The molar mass of an element or compound is the sum of the atomic masses of all its atoms.
grams of a substancemolar mass =
number of moles of the substancegrams of a monoatomic element
molar mass = number of moles of the element23
number of molecules number of moles =
6.022 x 10 molecules/mole
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Avogadro’s Number of Particles
6.022 x 1023 Particles
Molar Mass
1 MOLE
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1 mole = 6.022 x 1023 molecules1 mole = 6.022 x 1023 atoms1 mole = 6.022 x 1023 ions
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2 2Al + Fe2O3 Fe + Al2O3
• For calculations of mole-mass-volume relationships.
– The chemical equation must be balanced.
2 mol 2 mol1 mol 1 mol
The equation is balanced.
– The coefficient in front of a formula represents the number of moles of the reactant or product.
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Introduction to Stoichiometry:Introduction to Stoichiometry:The Mole-Ratio MethodThe Mole-Ratio Method
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• Stoichiometry: The area of chemistry that deals with the quantitative relationships between reactants and products.
• Mole Ratio: a ratio between the moles of any two substances involved in a chemical reaction.– The coefficients used in mole ratio
expressions are derived from the coefficients used in the balanced equation.
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2
23 m
1 mol
ol H
N
N2 + 3H2 2NH31 mol 2 mol3 mol
10
1 mol 2 mol3 molN2 + 3H2 2NH3
2
32 mo
3 mol H
l NH
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• The mole ratio is used to convert the number of moles of one substance to the corresponding number of moles of another substance in a stoichiometry problem.
• The mole ratio is used in the solution of every type of stoichiometry problem.
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The Mole Ratio Method
1. Convert the quantity of starting substance to moles (if it is not already moles)
2. Convert the moles of starting substance to moles of desired substance.
3. Convert the moles of desired substance to the units specified in the problem.
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Identify the starting substance from the data given in the problem statement. Convert the quantity of the starting substance to moles, if it is not already in moles.
1 molemoles = grams
molar mass
Step 1 Determine the number of moles of starting substance.
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1 mole NaClmoles NaCl = 292.15 grams NaCl = 5.000 moles NaCl
58.443 g NaCl
1 molemoles = grams
molar mass
How many moles of NaCl are present in 292.215 grams of NaCl? The molar mass of NaCl =58.443 g.
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1 mole NaClmoles NaCl = 292.15 grams NaCl = 5.0000 moles NaCl
58.443 g NaCl
1 molemoles = grams
molar mass
How many moles of NaCl are present in 292.215 grams of NaCl? The molar mass of NaCl =58.443 g.
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The number of moles of each substance in the balanced equation is indicated by the coefficient in front of each substance. Use these coefficients to set up the mole ratio.
moles of desired substance in the equationmole ratio =
moles of starting substance in the equation
Step 2 Determine the mole ratio of the desired substance to the starting substance.
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moles of desired substance
in the equationmoles of desired substance = moles of starting substance
moles of starting substance
in the equation
Step 2 Determine the mole ratio of the desired substance to the starting substance.
Multiply the number of moles of starting substance (from Step 1) by the mole ratio to obtain the number of moles of desired substance.
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moles of desired substance in the equationmoles of desired substance = moles of starting substance
moles of starting substance in the equation
In the following reaction how many moles of PbCl2
are formed if 5.000 moles of NaCl react?2NaCl(aq) + Pb(NO3)2(aq) PbCl2(s) + 2NaNO3(aq)
5.000 moles NaCl 2moles of PbCl = 22.500 mol PbCl21 mol PbCl
2 mol NaCl
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Step 3. Calculate the desired substance in the units specified in the problem.
• If the answer is to be in moles, the calculation is complete
• If units other than moles are wanted, multiply the moles of the desired substance (from Step 2) by the appropriate factor to convert moles to the units required.
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Step 3. Calculate the desired substance in the units specified in the problem.
molar mass1. To calculate : grams =gr moles x
1 moams
l
22 2
2
18.02 g H O90.10 grams H O = 5.000 mol H O
1 mol H
O
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236.022 x 10 atoms2. To calculate : atoms = moles
1 moatoms
l
2324 6.022 x 10 Na atoms
3.011 x 10 Na atoms = 5.000 moles Na atoms 1 mol Na
atoms
Step 3. Calculate the desired substance in the units specified in the problem.
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236.022 x 10 molecules3. To calculate : molecules = mol moles x
1 molecules
2324 2
2 22
6.022 x 10 H O molecules3.011 x 10 molecules H O = 5.000 moles H O
1 mol H O
Step 3. Calculate the desired substance in the units specified in the problem.
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Mole-Mole CalculationsMole-Mole Calculations
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Phosphoric Acid• Phosphoric acid (H3PO4) is one of the
most widely produced industrial chemicals in the world.
• Most of the world’s phosphoric acid is produced by the wet process which involves the reaction of phosphate rock, Ca5(PO4)3F, with sulfuric acid (H2SO4).
Ca5(PO4)3F(s) + 5H2SO4 3H3PO4 + HF + 5CaSO4
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Mole Ratio
Calculate the number of moles of phosphoric acid (H3PO4) formed by the reaction of 10 moles of sulfuric acid (H2SO4).
Ca5(PO4)3F + 5H2SO4 3H3PO4 + HF + 5CaSO4
Step 1 Moles starting substance: 10.0 mol H2SO4
Step 2 The conversion needed is moles H2SO4 moles H3PO4
1 mol 5 mol 3 mol 1 mol 5 mol
3 42 4
2 4
3 mol H PO10 mol H SO x =
5 mol H SO 3 46 mol H PO
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Step 2 The conversion needed is moles Ca5(PO4)3F moles H2SO4
Calculate the number of moles of sulfuric acid (H2SO4) that react when 10 moles of Ca5(PO4)3 react.
Ca5(PO4)3F + 5H2SO4 3H3PO4 + HF + 5CaSO4
Mole Ratio
Step 1 The starting substance is 10.0 mol Ca5(PO4)3F
1 mol 5 mol 3 mol 1 mol 5 mol
2 45 4 3
5 4 3
5 mol H SO10 mol Ca (PO ) F x =
1 mol Ca (PO ) F 2 450 mol H SO
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Mole-Mass CalculationsMole-Mass Calculations
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1. The object of this type of problem is to calculate the mass of one substance that reacts with or is produced from a given number of moles of another substance in a chemical reaction.
2. If the mass of the starting substance is given, we need to convert it to moles.
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3. We use the mole ratio to convert moles of starting substance to moles of desired substance.
4. We can then change moles of desired substance to mass of desired substance if called for by the problem.
30Mole Ratio
2 4
3 4
5 mol H SO =
3 mol H PO
Calculate the number of moles of H2SO4 necessary to yield 784 g of H3PO4.
Ca5(PO4)3F+ 5H2SO4 3H3PO4 + HF + 5CaSO4
Method 1 Step by Step
Step 1 The starting substance is 784 grams of H3PO4.
Step 2 Convert grams of H3PO4 to moles of H3PO4.
Step 3 Convert moles of H3PO4 to moles of H2SO4 by the mole-ratio method.
3 4
3 4
1 mol H PO =
98.0 g H PO
3 48.00 mol H PO
2 413.3 mol H SO
3 4784 g H PO
3 48.00 mol H PO
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Mole Ratio
2 4
3 4
5 mol H SO =
3mol H PO
Calculate the number of moles of H2SO4 necessary to yield 784 g of H3PO4
Ca5(PO4)3F+ 5H2SO4 3H3PO4 + HF + 5CaSO4
Method 2 Continuous
grams H3PO4 moles H3PO4 moles H2SO4
The conversion needed is
3 4784 g H PO 3 4
3 4
1 mol H PO
98.0 g H PO
2 413.3 mol H SO
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218.0 mol H 2
2
2.02 g H =
1 mol H
Method 1 Step by Step
Step 1 The starting substance is 12.0 moles of NH3
Step 2 Calculate moles of H2 by the mole-ratio method.
Step 3 Convert moles of H2 to grams of H2.
Calculate the number of grams of H2 required to form 12.0 moles of NH3.
N2 + 3H2 2NH3
312.0 mol NH 218.0 mol H
Mole Ratio
2
3
3 mol H =
2 mol NH
236.0 g H
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Mole Ratio
2
3
3 mol H
2 mol NH
moles NH3 moles H2 grams H2
Calculate the number of grams of H2 required to form 12.0 moles of NH3.
N2 + 3H2 2NH3
Method 2 Continuous
The conversion needed is
312.0 mol NH 2
2
2.02 g H =
1mol H
236.0 g H
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Mass-Mass CalculationsMass-Mass Calculations
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Solving mass-mass stoichiometry problems requires all the steps of the mole-ratio method.
1. The mass of starting substance is converted to moles.
2. The mole ratio is then used to determine moles of desired substance.
3. The moles of desired substance are converted to mass of desired substance.
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Calculate the number of grams of NH3 formed by the reaction of 112 grams of H2.
N2 + 3H2 2NH3Method 1 Step by Step
Step 1 The starting substance is 112 grams of H2. Convert 112 g of H2 to moles.
grams moles2
2
1 mol H
2.02 g H
2112 g H 255.4 moles H
Step 2 Calculate the moles of NH3 by the mole ratio method.
3
2
2 mol NH=
3 mol H
255.4 moles H 336.9 moles NH
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Calculate the number of grams of NH3 formed by the reaction of 112 grams of H2.
N2 + 3H2 2NH3Method 1 Step by Step
Step 3 Convert moles NH3 to grams NH3.
moles grams
336.9 moles NH 3
3
17.0 g NH=
1 mol NH
3629 g NH
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Calculate the number of grams of NH3 formed by the reaction of 112 grams of H2.
N2 + 3H2 2NH3
grams H2 moles H2 moles NH3 grams NH3
2
2
1 mol H
2.02 g H
2112 g H 3
2
2 mol NH
3 mol H
Method 2 Continuous
3
3
17.0 g NH=
1 mol NH
3629 g NH
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Limiting-Reactant and Limiting-Reactant and Yield CalculationsYield Calculations
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• It is called the limiting reactant because the amount of it present is insufficient to react with the amounts of other reactants that are present.
• The limiting reactant limits the amount of product that can be formed.
• The limiting reactant is one of the reactants in a chemical reaction.
Limiting ReactantLimiting ReactantLimiting ReactantLimiting Reactant
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How many bicycles can be assembled from the parts shown?
From eight wheels four bikes can be constructed.
From four frames four bikes can be constructed.
From three pedal assemblies three bikes can be constructed.
The limiting part is the number of pedal assemblies.
9.2
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H2 + Cl2 2HCl
+
7 molecules H2 can form 14 molecules HCl
4 molecules Cl2 can form 8 molecules HCl 3 molecules of H2 remain
H2 is in excess Cl2 is the limiting
reactant 9.3
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1. Calculate the amount of product (moles or grams, as needed) formed from each reactant.
2. Determine which reactant is limiting. (The reactant that gives the least amount of product is the limiting reactant; the other reactant is in excess.
3. Calculate the amount of the other reactant required to react with the limiting reactant, then subtract this amount from the starting quantity of the reactant. This gives the amount of the that substance that remains unreacted.
Steps Used to Determine the Limiting Steps Used to Determine the Limiting ReactantReactant
Steps Used to Determine the Limiting Steps Used to Determine the Limiting ReactantReactant
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How many moles of HCl can be produced by reacting 4.0 mol H2 and 3.5 mol Cl2? Which compound is the limiting reactant?
Step 1 Calculate the moles of HCl that can form from each reactant.
24.0 mol H2
2 mol HCl
1 mol H
8.0 mol HCl
23.5 mol Cl2
2 mol HCl
1 mol Cl
7.0 mol HCl
H2 + Cl2 → 2HCl
Step 2 Determine the limiting reactant.
The limiting reactant is Cl2 because it produces less HCl than H2.
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How many moles of silver bromide (AgBr) can be formed when solutions containing 50.0 g of MgBr2 and 100.0 g of AgNO3 are mixed together? How many grams of the excess reactant remain unreacted?
Step 1 Calculate the grams of AgBr that can form from each reactant.
MgBr2(aq) + 2AgNO3 (aq) → 2AgBr(s) + Mg(NO3)2(aq)
The conversion needed isg reactant → mol reactant → mol AgBr → g AgBr 250.0 g MgBr 102 g AgBr2
2
1 mol MgBr
184.1 g MgBr
2
2 mol AgBr
1 mol MgBr
187.8 g AgBr
1 mol AgBr
3100.0 g AgNO 110.5 g AgBr3
3
1 mol AgNO
169.9 g AgNO
3
2 mol AgBr
2 mol AgNO
187.8 g AgBr
1 mol AgBr
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How many moles of silver bromide (AgBr) can be formed when solutions containing 50.0 g of MgBr2 and 100.0 g of AgNO3 are mixed together? How many grams of the excess reactant remain unreacted?
Step 2 Determine the limiting reactant.
MgBr2(aq) + 2AgNO3 (aq) → 2AgBr(s) + Mg(NO3)2(aq)
250.0 g MgBr 102 g AgBr2
2
1 mol MgBr
184.1 g MgBr
2
2 mol AgBr
1 mol MgBr
187.8 g AgBr
1 mol AgBr
3100.0 g AgNO 110.5 g AgBr3
3
1 mol AgNO
169.9 g AgNO
3
2 mol AgBr
2 mol AgNO
187.8 g AgBr
1 mol AgBr
The limiting reactant is MgBr2 because itforms less Ag Br.
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How many grams of the excess reactant (AgNO3) remain unreacted?
Step 3 Calculate the grams of unreacted AgNO3. First calculate the number of grams of AgNO3 that will react with 50 g of MgBr2.
MgBr2(aq) + 2AgNO3 (aq) → 2AgBr(s) + Mg(NO3)2(aq)
250.0 g MgBr 392.3 g AgNO2
2
1 mol MgBr
184.1 g MgBr
3
2
2 mol AgNO
1 mol MgBr
3
3
169.9 g AgNO
1 mol AgNO
The conversion needed isg MgBr2 → mol MgBr2 → mol AgNO3 → g AgNO3
The amount of MgBr2 that remains is
100.0 g AgNO3 - 92.3 g AgNO3 = 7.7 g AgNO3
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The quantities of products calculated from equations represent the maximum yield (100%) of product according to the reaction represented by the equation.
Reaction YieldReaction YieldReaction YieldReaction Yield
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Many reactions fail to give a 100% yield of product.
This occurs because of side reactions and the fact that many reactions are reversible.
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• The theoretical yield of a reaction is the calculated amount of product that can be obtained from a given amount of reactant.
• The actual yield is the amount of product finally obtained from a given amount of reactant.
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• The percent yield of a reaction is the ratio of the actual yield to the theoretical yield multiplied by 100.
actual yield x 100 = percent yield
theoretical yield
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187.8 g AgBr
1 mol AgBr
Silver bromide was prepared by reacting 200.0 g of magnesium bromide and an adequate amount of silver nitrate. Calculate the percent yield if 375.0 g of silver bromide was obtained from the reaction:
MgBr2(aq) + 2AgNO3 (aq) → 2AgBr(s) + Mg(NO3)2(aq)
Step 1 Determine the theoretical yield by calculating the grams of AgBr that can be formed.
The conversion needed isg MgBr2 → mol MgBr2 → mol AgBr → g AgBr
2200.0 g MgBr 408.0 g AgBr2
2
1 mol MgBr
184.1 g MgBr
2
2 mol AgBr
1 mol MgBr
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Silver bromide was prepared by reacting 200.0 g of magnesium bromide and an adequate amount of silver nitrate. Calculate the percent yield if 375.0 g of silver bromide was obtained from the reaction:
MgBr2(aq) + 2AgNO3 (aq) → 2AgBr(s) + Mg(NO3)2(aq)
Step 2 Calculate the percent yield.
actual yieldpercent yield = x 100
theoretical yield
percent yield = 375.0 g AgBr
x 100 =408.0 g AgBr
91.9%
must have same units
must have same units
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Key Concepts9.1 A Short Review
9.2 Introduction to Stoichiometry:The Mole-Ratio Method
9.3 Mole-Mole Calculations
9.4 Mole-Mass Calculations
9.5 Mass-Mass Calculations
9.6 Limiting-Reactant and Yield Calculations