1 Calculations From Chemical Equations Chapter 9 Hein and Arena Eugene Passer Chemistry Department...

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Calculations FromChemical Equations

Chapter 9

Calculations FromChemical Equations

Chapter 9

Hein and Arena

Eugene PasserChemistry DepartmentBronx Community College

© John Wiley and Sons, Inc.

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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A Short ReviewA Short Review

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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.

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The molar mass of an element or compound is the sum of the atomic masses of all its atoms.

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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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ExamplesExamplesExamplesExamples

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2

23 m

1 mol

ol H

N

N2 + 3H2 2NH31 mol 2 mol3 mol

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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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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


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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


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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.

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35Mole 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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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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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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Limiting ReactantLimiting ReactantLimiting ReactantLimiting Reactant

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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.

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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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Steps Used to Determine Steps Used to Determine the Limiting Reactantthe Limiting Reactant

Steps Used to Determine Steps Used to Determine the Limiting Reactantthe Limiting Reactant

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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.

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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.


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.


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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Reaction YieldReaction YieldReaction YieldReaction Yield

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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.

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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:


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:


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

1 Calculations From Chemical Equations Chapter 9 Hein and Arena Eugene Passer Chemistry Department...

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