Ch9-Stoichiometry 2

7/27/2019 Ch9-Stoichiometry 2

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1

Stoichiometry

Chapter 9

StoichiometryChapter 9

Version 1.0


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Objectives

Define stoichiometry

Solve stoichiometric problems involving

mole-ratio method

Mass-mass relation

Mass-volume relation

Volume-volume relation

Define excess reagent, limiting reagent, and

percent yield

Differentiate theoretical from actual yield 2


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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

(Avogadros number) of the element.


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


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Formula Weight

Formula weightFormula weight: the sum of the atomicweights in atomic mass units (amu) of all

atoms in a compounds formula:

Ionic Compounds

Sodium chloride (NaCl) 23.0 amu Na + 35.5 amu Cl = 58.5 amu

Aspirin (C9H8O4) 9(12.0 amu C) + 8(1.0 amu H) +4(16.0 amu O) = 180.0 amu

Water (H2O) 2(1.0 amu H) + 16.0 amu O = 18.0 amu

Nickel(II) chloride hydrate(NiCl26H2O)

58.7 amu Ni + 2(35.5 amu Cl) +12(1.0) amu H) + 6(12.0 amu O) = 237.7 amu

Molecular Compounds


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Formula Weight

Formula weightFormula weight can be used for both ionicand molecular compounds; it tells nothing

about whether a compound is ionic or

molecular. Molecular weightMolecular weight should be used only for

molecular compounds.

For comparison, we use formula weight forionic compounds and molecular weight for

molecular compounds.


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grams of a substancemolar mass =number of moles of the substance

grams of a monoatomic elementmolar mass =number of moles of the element

23number of moleculesnumber of moles =

6.022 x 10 molecules/mole


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Avogadros

Number ofParticles

6 x 1023

Particles

Molar Mass

1 MOLE


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2 2Al + Fe2O3 Fe + Al2O3

For calculations of mole-mass-volume

relationships.

The chemical equation mustbe balanced.

2 mol 2 mol1 mol 1 mol

The equation is balanced.

The number in front of a formula

represents the number of moles of the

reactant or product.


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Stoichiometry: The area of chemistry

that deals with the quantitativerelationships between reactants and

products.

Mole Ratio: a ratio between the molesof any two substances involved in a

chemical reaction.

The coefficients used in mole ratioexpressions are derived from the

coefficients used in the balanced

equation.


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Stoichiometry

Stoichiometry:Stoichiometry: the study of mass relationshipsin chemical reactions

following is an overview of the types of

calculations we study

M o l e s o f AG r a m s o f AG r a m s o f BM o l e s o f B

F r o m m o l e s t o m o l e s ,u s e t h e c o e f f i c i e n t s i n

t h e b a l a n c e d e q u a t i o na s a c o n v e r s i o n f a c t o r

F r o m g r a m s t o m o l e s ,u s e m o l a r m a s s ( g / m o l )

a s a c o n v e r s i o n f a c t o r

F r o m m o l e s t o g r a m s ,u s e m o l a r m a s s ( g / m o l )

a s a c o n v e r s i o n f a c t o r

Y o u a r e g i v e n o n e o f t h e s e A n d a s k e d t o f i n d o n e o f t h e s e


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ExamplesExamplesExamplesExamples


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2

23 m

1 mol

olH

N

N2 + 3H2 2NH31 mol 2 mol3 mol


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1 mol 2 mol3 mol

N2 + 3H2 2NH3

2

32 mo

3 molH

l NH


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The mole ratio is used to convert thenumber of moles of one substance to

the corresponding number of moles of

another substance in a stoichiometryproblem.

The mole ratio is used in the solution of

every type of stoichiometry problem.


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Stoichiometry

Problem: how many grams of nitrogen, N2, arerequired to produce 7.50 g of ammonia, NH3

first find how many moles of NH3 are in7.50 g of NH3

next find how many moles of N2 arerequired to produce this many moles of NH3

7.50 g NH3 x1 mol NH3

17.0 g NH3= mol NH3

7.50 g NH3 x1 mol NH3

17.0 g NH3

x1 mol N2

2 mol NH3= mol N2

N2 (g) + 2NH3 ( g)3H2 (g)


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Stoichiometry

Finally convert moles of N2 to grams of N2and now do the math

x28.0 g N2

1 mol N2= 6.18 g N27.50 g NH3 x

1 mol NH3

17.0 g NH3

x1 mol N2

2 mol NH3


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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

done.1 mole

moles = gramsmolar mass

Step 1. Determine the number of moles of starting

substance.


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The number of moles of each substance in the

balanced equation is indicated by thecoefficient 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.

h f ll i i h l f bCl


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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 PbCl2are formed if 5.000 moles of NaCl react?

2NaCl(aq) + Pb(NO3)2(aq) PbCl2(s) + 2NaNO3(aq)

5.000 moles NaCl2moles of PbCl = 22.500 mol PbCl2

1 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 atoms

2. To calculate : atoms = moles1 mo

atomsl

2324 6.022 x 10 Na atoms3.011 x 10 Na atoms = 5.000 moles Na atoms

1 mol Na atoms

Na




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236.022 x 10 molecules

3. To calculate : molecules =mol moles x1 mol

ecules

2324 2

2 2

2

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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Stoichiometry

Problem 1. What mass of aluminum oxide isrequired to prepare 27 g of aluminum?

Problem 2. How many grams each of CO22

and NH33

are produced from 0.83 mol of

urea?

Al2O3 ( s)electrolysis Al(s) + O2 ( g)

(NH2 )2CO(aq) + 2NH3 (aq) + CO2 (g)H2O

Urea

urease


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Mole-MoleMole-MoleCalculationsCalculations


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Phosphoric Acid

Phosphoric acid (H3PO4) is one of the most

widely produced industrial chemicals in the

world.

Most of the worlds phosphoric acid is

produced by the wet process which involves

the reaction of phosphate rock, Ca5(PO4)3,F

with sulfuric acid (H2SO4).

Ca5(PO4)3F(s) + 5H2SO4 3H3PO4 + HF + 5CaSO4

C l l t th b f l f h h i id


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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 SO3 4

6 mol H PO

C l l t th b f l f lf i id


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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 with 10 moles of Ca5(PO4)3F.

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 ) F2 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 onesubstance that reacts with or is

produced from a given number of

moles of another substance in achemical 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 molesof desired substance.

4. We can then change moles of desiredsubstance to mass of desired

substance if called for by the

problem.


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Calculate the number of moles of H SO necessary to


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38Mole 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 + 5CaSO4Method 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 themole-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

Calculate the number of moles of H SO necessary to


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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 + 5CaSO4Method 2 Continuous

grams H3PO4 moles H3PO4 moles H2SO4

The conversion needed is

3 4

784 g H PO 3 4

3 4

1 mol H PO

98.0 g H PO

2 413.3 mol H SO

Calculate the number of grams of H required to form


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2

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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

2

1

8.0

Mole Ratio

2

3

3 mol H=

2 mol NH

22

2.0

2g

1m

ol

=

Calculate the number of grams of H required to form


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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

2NH3Method 2 Continuous


312.0 mol NH2

2

2.02 gH=

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-ratiomethod.

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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Mass-mass Relations

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g

u

g

ugu

NN

MmMmmm =

mu = mass of unknown substancemg = mass of given substance

Mmu = molar mass of the given substance

Mmg = molar mass of the given substance

Nu = coefficient of the unknown in the balanced

equation

Ng = coefficient of the given substance in the

balanced equation

Calculate the number of grams of NH formed by the


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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 gH

=

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 NH3

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 H2.02 g H

2112 gH 32

2 mol NH3 mol H

Method 2 Continuous

3

3

17.0 g NH =1 mol NH

3629 g NH


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Mass-VolumeMass-VolumeCalculationsCalculations


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Mass-volume Relations

Remember:1 mole of any substance occupies 22.4 L

Gases are assumed to behave as ideal

gasesThe reaction must take place at STP

(standard temperature and pressure)

where T = 0o C = 273 KP = 1 atm = 760 mm Hg

As gas not in STP must be converted to49


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Mole-Volume CalculationsMole-Volume Calculations

Mass-Volume CalculationsMass-Volume Calculations


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giveng

u

g

gumol1

L4.22

N

N

Mm

1mvol =

L22.4mol1

NNMmvm given

g

uugu =

For unknown volume

For unknown mass


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volu = volume of unknown substance

mu = mass of unknown substance

mg = given mass

vg = given volumeMmg = molar mass of the given substance

Mmu = molar mass of the unknown substance

Nu

= coefficient of the unknown in the balanced

equation

Ng = coefficient of the given substance in the

balanced equation

What volume of oxygen (at STP) can be formed


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Step 1 Write the balanced equation

2 KClO3 2 KCl + 3 O2

Step 2 The starting amount is 0.500

mol KClO3. The conversion is

moles KClO3 moles O2 liters O2


from 0.500 mol of potassium chlorate?



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2

3

3 mol O

2 mol KClO

Step 3. Calculate the moles of O2, using the

mole-ratio method.

3(0.500 mol KClO )


from 0.500 mol of potassium chlorate?

Step 4. Convert moles of O2 to liters of O2

2= 0.750 mol O

2(0.750 mol O )22.4 L

1 mol

2= 16.8 L O

2 KClO3 2KCl + 3 O2


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Volume-VolumeVolume-VolumeCalculationsCalculations


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Volume-volume Relations

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g

ugu

N

Nvv =

This can be solved by ratio and proportion

vu= volume of unknown substance

vg= volume of given substance

Nu

= number of moles of unknown substance from

the balanced equation

ng = number of moles of given substance from the

balanced equation

For reacting gases at constant temperature and


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H2(g) + Cl2(g) 2HCl(g)

1 mol H2 1 mol Cl2 2 mol HCl22.4 L

STP

22.4 L

STP

2 x 22.4 L

STP

1 volume 1 volume 2 volumes

Y volume Y volume 2Y volumes

g g p

pressure: Volume-volume relationships are the same

as mole-mole relationships.

What volume of oxygen will react with 150 L of


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yg

hydrogen to form water vapor? What volume of water

vapor will be formed?

2H2(g) + O2(g)

2H2O(g)

Assume that both reactants and products are measured at

STP. Calculate by using reacting volumes:

2 mol 1 mol 2 mol

2 x 22.4 L 22.4 L 2 x 22.4 L

2 volumes 1 volume 2 volumes

What volume of oxygen will react with 150 L of


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yg

hydrogen to form water vapor? What volume of water

vapor will be formed?

2H2(g) + O2(g) 2H2O(g)

( ) 22

22 OL75

Hvol2Ovol1HL150 =

( ) OHL150Hvol2

OHvol2HL150 2

2

22 =

What volume of nitrogen will react with 600. mL of


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g

hydrogen to form ammonia? What volume of

ammonia will be formed?

N2(g) + 3H2(g) 2NH3(g)

2600. ml H 2

2

1 vol N3 vol H

2= 200. mL N

2600. ml H3

2

2 vol NH3 vol H

3= 400. mL NH


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Limiting-Reactant andLimiting-Reactant andYield CalculationsYield Calculations


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Limiting Reactant/ReagentLimiting Reactant/ReagentLimiting Reactant/ReagentLimiting Reactant/Reagent


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It is called the limiting reactant

because the amount of it present isinsufficient to react with the amounts

of other reactants that are present.

The limiting reactant limits the amountof product that can be formed.

The limiting reactant/reagent is one of

the reactants in a chemical reaction.

How many bicyclesFrom eight wheels four

From four frames fourFrom three pedal assemblies


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can be assembled

from the parts

shown?

bikes can be constructed.bikes can be constructed.three bikes can be constructed.

The limiting part is thenumber of pedal

assemblies.

9.2

H + Cl 2HCl


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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

reactant9.3


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Steps Used to Determine theSteps Used to Determine theLimiting ReactantLimiting ReactantSteps Used to Determine theSteps Used to Determine theLimiting ReactantLimiting Reactant

1 Calculate the amount of product (moles or


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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. (Thereactant 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 startingquantity of the reactant. This gives the

amount of the substance that remains

unreacted.


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How many moles of HCl can be produced by reacting


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4.0 mol H2 and 3.5 mol Cl2? Which compound is the

limiting reagent?

Step 1 Calculate the moles of HCl that can form

from each reactant.

24.0 mol H

2

2 mol HCl1 mol H

=

8.0 mol HCl

23.5 mol Cl

2

2 mol HCl

1 mol Cl

=

7.0 mol HCl

H2 + Cl2 2HCl

Step 2 Determine the limiting reactant.

The limiting reactant is Cl2because it

producesless HCl than H2.

How many grams of silver bromide (AgBr) can be


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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)


g reactant mol reactant mol AgBr g

AgBr( )250.0 g MgBr 102 g AgBr

2

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

=

How many grams of silver bromide (AgBr) can be


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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 it

forms less Ag Br.

How many grams of the excess reactant (AgNO3)


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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 AgNO22

1 mol MgBr

184.1 g MgBr

3

2

2 mol AgNO

1 mol MgBr

3

3

169.9 g AgNO

1 mol AgNO

=


g 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 maximumyield (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 isthe 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 theratio of the actual yield to the theoretical

yield multiplied by 100.

actual yieldx 100 = percent yield

theoretical yield

Silver bromide was prepared by reacting 200.0 g of

i b id d d f il


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187.8 g AgBr

1 mol AgBr

=

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 beformed.


g 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

Silver bromide was prepared by reacting 200.0 g of

i b id d d f il


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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

Solve the following problems correctly.


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g p y

1. Given the equation

Al4C3 + 12 H2O 4 Al(OH)3 + 3 CH4

a. How many moles of water are needed to react

with 100 g of Al4C3 ?

b. How many moles of Al(OH)3 will be produced

when 0.600 mol of CH4 is formed?

2. How many grams of zinc phosphate are formed

when 10.0 g of Zn are reacted with phosphoric

acid?

3 Zn + 2 H3PO4 Zn3(PO4)2 + 3 H2 83

Solve the following problems correctly.1 Gi th ti


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1. Given the equation

4 FeS2 + 11 O2 2 Fe2O3 + 8 SO2

a. How many moles ofFe2O3 can be made from 1.00 mol

of FeS2?

b. How many moles of O2

are required to react with

4.5 mol of FeS2?

c. If the reaction produces 1.55 mol ofFe2O3 , how

many mol of SO2

are produced?

d. How many grams of SO2 can be formed from 0.512

mol of FeS2?

e. How many grams of FeS2 are needed to produce 22184

In the following equations, determine which reactant


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is the limiting reactant and which reactant is in excess.

1. 2 Bi(NO3)3 + 3 H2S Bi2S3 + 6 HNO3

50.0 g 6.00 g

2. 3 Fe + 4 H2O Fe3O4 + 4 H2

40.0 g 16.0 g

85

In the following equations, determine which reactant


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is the limiting reactant and which reactant is in excess.

Acetylene (C2H2) can be manufactured bythe reaction of water and calcium carbide,

CaC2:

CaC2 + 2 H2O C2H2 + Ca(OH)2

When 44.5 g of commercial grade (impure)

calcium carbide is reacted, 0.540 mol of C2H2, isproduced. Assuming that all CaC2 was reacted to

C2H2 , what is the percent of CaC2 in the

commercial grade material?86


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Documents

Ch9-Stoichiometry 2