1 Lecture 4 Chemical Reactions and Quantities Chemical Reactions and Equations Copyright © 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings

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Lecture 4 Chemical Reactions and Quantities

Chemical Reactions and Equations

Copyright © 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings

Post-exam Error Analysis

Work in groups of 3 on exam Maximum of 15 points possible

15 – (1/5)(theoretical – actual) This is what I call PEA score

Exam Correction Factor (wrong answer points / right answer points)

Points you earned: Correction factor x PEA score Examples on the board

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Chemical Reaction

In a chemical reaction, A chemical change

produces one or more new substances.

There is a change in the composition of one or more substances.

Old bonds are broken and new ones are formed


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Chemical Equations

A chemical equation

Gives the chemical formulas of the reactants on the left of the arrow and the products on the right.

Reactants Product

C(s)

O2 (g)CO2 (g)


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Symbols Used in Equations

Symbols are used in

chemical equations

to show The states of the

reactants.

The states of the products.

The reaction conditions.

TABLE 5.2


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Chemical Equations are Balanced


Chemical equations must be balanced!

Atoms are not gained or lost.

The number of atoms in the reactants is equal to the number of atoms in the products.

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A Balanced Chemical Equation

In a balanced chemical equation, There must be the same number of each type of

atom on the reactant side and on the product side.

Al + S Al2S3 Not Balanced

2Al + 3S Al2S3 Balanced

2Al = 2Al

3S = 3S

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Learning Check

State the number of atoms of each element on the

reactant side and the product side for each of the

following balanced equations:

A. P4(s) + 6Br2(l) → 4 PBr3(g)

B. 2Al(s) + Fe2O3(s) → 2Fe(s) + Al2O3(s)

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Learning Check

Determine if each equation is balanced or not.

A. Na(s) + N2(g) → Na3N(s)

B. C2H4(g) + H2O(l) → C2H5OH(l)

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Checking a Balanced Equation

Reactants Products

1 C atom = 1 C atom

4 H atoms = 4 H atoms

4 O atoms = 4 O atoms


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Guide to Balancing Equations

Write out correct formulas Count the atoms of each element on both

sides of the equation Never change subscripts Use coefficients to balance each element Do oxygen last Hydrogen is next to last Check final equation for balance

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Balance the following chemical equation:

NH3(g) + O2(g) NO(g) + H2O(g)

Balancing Chemical Equations

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Check the balance of atoms in the following:

Fe3O4(s) + 4H2(g) 3Fe(s) + 4H2O(l)

A. Number of H atoms in products.

1) 2 2) 4 3) 8

B. Number of O atoms in reactants.

1) 2 2) 4 3) 8

C. Number of Fe atoms in reactants.

1) 1 2) 3 3) 4

Learning Check

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Balance each equation and list the coefficients in the balanced equation going from reactants to products:

A. __Mg(s) + _N2(g) __Mg3N2(s)

1) 1, 3, 2 2) 3, 1, 2 3) 3, 1, 1

B. __Al(s) + __Cl2(g) __AlCl3(s)

1) 3, 3, 2 2) 1, 3, 1 3) 2, 3, 2

Learning Check

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Equations with Polyatomic Ions


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Learning Check

Barium nitrate reacts with sodium carbonate to form sodium nitrate and barium carbonate. What is the balanced chemical equation?

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Balance and list the coefficients from reactants to products:

A. __Fe2O3(s) + __C(s) __Fe(s) + __CO2(g)

1) 2, 3, 2,3 2) 2, 3, 4, 3 3) 1, 1, 2, 3

B. __Al(s) + __FeO(s) __Fe(s) + __Al2O3(s)

1) 2, 3, 3, 1 2) 2, 1, 1, 1 3) 3, 3, 3, 1

C. __Al(s) + __H2SO4(aq) __Al2(SO4)3(aq) + __H2(g)

1) 3, 2, 1, 2 2) 2, 3, 1, 3 3) 2, 3, 2, 3

Learning Check

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Lecture 4.0: Chemical Reactions and Quantities

Types of Reactions


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Type of Reactions

Chemical reactions can be classified as Combination reactions.

Decomposition reactions.

Single Replacement reactions.

Double Replacement reactions.

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Combination

In a combination reaction, Two or more elements (or simple compounds)

combine to form one product

+

2Mg(s) + O2(g) 2MgO(s)

2Na(s) + Cl2(g) 2NaCl(s)

SO3(g) + H2O(l) H2SO4(aq)

A B A B

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Decomposition

In a decomposition reaction, One substance splits into two or more simpler

substances.

2HgO(s) 2Hg(l) + O2(g)

2KClO3(s) 2KCl(s) + 3O2(g)


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Single Replacement

In a single replacement reaction, One element takes the place of a different element in

a reacting compound.

Zn(s) + 2HCl(aq) ZnCl2(aq) + H2(g)

Fe(s) + CuSO4(aq) FeSO4(aq) + Cu(s)


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Zn and HCl is a Single Replacement Reaction


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Double Replacement

In a double replacement, Two elements in the reactants exchange places.

AgNO3(aq) + NaCl(aq) AgCl(s) + NaNO3(aq)

ZnS(s) + 2HCl(aq) ZnCl2(aq) + H2S(g)


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Example of a Double Replacement


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Chemical Reactions and Quantities

Oxidation-Reduction Reactions


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Oxidation and Reduction

An oxidation-reduction reaction Provides us with energy from food. Provides electrical energy in batteries. Occurs when iron rusts.

4Fe(s) + 3O2(g) 2Fe2O3(s)Copyright © 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings

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An oxidation-reduction reaction

Transfers electrons from one reactant to another.

A Loss of Electrons is Oxidation (LEO)

Zn(s) Zn2+(aq) + 2e-

A Gain of Electrons is Reduction (GER)Cu2+(aq) + 2e- Cu(s)

Electron Loss and Gain

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Oxidation and Reduction


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Zn and Cu2+

oxidationZn(s) Zn2+(aq) + 2e-Silvery metal

reduction Cu2+(aq) + 2e- Cu(s)

Blue orange Copyright © 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings

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Electron Transfer from Zn to Cu2+

Oxidation: electron loss

Reduction: electron gain


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Write the separate oxidation and reduction reactions for the following equation.

2Cs(s) + F2(g) 2CsF(s)

Each cesium atom loses an electron to form cesium ion.

2Cs(s) 2Cs+(s) + 2e− oxidation

Fluorine atoms gain electrons to form fluoride ions.

F2(s) + 2e- 2F−(s) reduction

Writing Oxidation and Reduction Reactions

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Pause for ALE exercise

Do, in groups, problems from the active learning exercise sheet

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The Mole


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Collection Terms

A collection term states a specific number of items.

1 dozen donuts = 12 donuts

1 ream of paper = 500 sheets

1 case = 24 cans


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A mole is a collection that contains The same number of particles as there are

carbon atoms in 12.0 g of carbon 12C. 6.02 x 1023 atoms of an element (Avogadro’s

number).

1 mole element Number of Atoms

1 mole C = 6.02 x 1023 C atoms

1 mole Na = 6.02 x 1023 Na atoms

1 mole Au = 6.02 x 1023 Au atoms

A Mole of Atoms

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A mole Of a covalent compound has Avogadro’s number of

molecules.

1 mole CO2 = 6.02 x 1023 CO2 molecules

1 mole H2O = 6.02 x 1023 H2O molecules

Of an ionic compound contains Avogadro’s number of formula units.

1 mole NaCl = 6.02 x 1023 NaCl formula units

1 mole K2SO4 = 6.02 x 1023 K2SO4 formula units

A Mole of a Compound

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Particle in One-Mole Samples

TABLE 5.3

Copyright © 2007 by Pearson Education, Inc Publishing as Benjamin Cummings

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Avogadro’s number 6.02 x 1023 can be written as an

equality and two conversion factors.

Equality:

1 mole = 6.02 x 1023 particles

Conversion Factors:

6.02 x 1023 particles and 1 mole 1 mole 6.02 x 1023 particles

Avogadro’s Number

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Using Avogadro’s Number

Avogadro’s number is used to convertmoles of a substance to particles.

How many Cu atoms are in

0.50 mole Cu?

0.50 mole Cu x 6.02 x 1023 Cu atoms 1 mole Cu

= 3.0 x 1023 Cu atoms


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Using Avogadro’s Number

Avogadro’s number is used to convertparticles of a substance to moles.

How many moles of CO2 are in

2.50 x 1024 molecules CO2?

2.50 x 1024 molecules CO2 x 1 mole CO2

6.02 x 1023 molecules CO2

= 4.15 moles CO2

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Subscripts and Moles

The subscripts in a formula give The relationship of atoms in the formula. The moles of each element in 1 mole of compound.

Glucose

C6H12O6

In 1 molecule: 6 atoms C 12 atoms H 6 atoms O

In 1 mole: 6 moles C 12 moles H 6 moles O

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Subscripts State Atoms and Moles

1 mole C9H8O4 = 9 moles C 8 moles H 4 moles O


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Factors from Subscripts

Subscripts used for conversion factors Relate moles of each element in 1 mole compound. For aspirin C9H8O4 can be written as:

9 moles C 8 moles H 4 moles O

1 mole C9H8O4 1 mole C9H8O4 1 mole C9H8O4

and

1 mole C9H8O4 1 mole C9H8O4 1 mole C9H8O4 9 moles C 8 moles H 4 moles O

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Molar Mass


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The molar mass is The mass of one mole of a substance. The atomic mass of an element expressed in grams.

Molar Mass


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Molar Mass of CaCl2

For a compound, the molar mass is the sum of the molar masses of the elements in the formula. We calculate the molar mass of CaCl2 to the nearest 0.1 g as follows.

Element Number of Moles

Atomic Mass Total Mass

Ca 1 40.1 g/mole 40.1 g

Cl 2 35.5 g/mole 71.0 g

CaCl2 111.1 g

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Molar Mass of K3PO4

Determine the molar mass of K3PO4 to 0.1 g.

Element Number of Moles

Atomic Mass Total Mass in K3PO4

K 3 39.1 g/mole 117.3 g

P 1 31.0 g/mole 31.0 g

O 4 16.0 g/mole 64.0 g

K3PO4 212.3 g

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One-Mole Quantities

32.1 g 55.9 g 58.5 g 294.2 g 342.3 g


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Methane CH4 known as natural gas is used in gas cook tops and gas heaters.

1 mole CH4 = 16.0 g

The molar mass of methane can be written as conversion factors.

16.0 g CH4 and 1 mole CH4

1 mole CH4 16.0 g CH4

Conversion Factors from Molar Mass

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Mole factors are used to convert between the grams of

a substance and the number of moles.

Calculations Using Molar Mass

Grams Mole factor Moles


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Aluminum is often used for the structure of

lightweight bicycle frames. How many grams

of Al are in 3.00 moles of Al?

3.00 moles Al x 27.0 g Al = 81.0 g Al1 mole Al

mole factor for Al

Calculating Grams from Moles

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Lecture 4.0 Chemical Reactions and Quantities

Mole Relationships in Chemical Equations


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Law of Conservation of Mass

The Law of Conservation of Mass indicates that in an

ordinary chemical reaction, Matter cannot be created or destroyed. No change in total mass occurs in a reaction. Mass of products is equal to mass of

reactants.

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Conservation of Mass

+

Reactants Products2 moles Ag + 1 moles S = 1 mole Ag2S2 (107.9 g) + 1(32.1 g) = 1 (247.9 g)

247.9 = 247.9 g


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Consider the following equation:4Fe(s) + 3O2(g) 2Fe2O3(s)

This equation can be read in “moles” by placing theword “moles” between each coefficient and formula.

4 moles Fe + 3 moles O2 2 moles Fe2O3

Reading Equations with Moles

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A mole-mole factor is a ratio of the moles for any two

substances in an equation.

4Fe(s) + 3O2(g) 2Fe2O3(s)

Fe and O2 4 moles Fe and 3 moles O2

3 moles O2 4 moles Fe

Fe and Fe2O3 4 moles Fe and 2 moles Fe2O3

2 moles Fe2O3 4 moles Fe

O2 and Fe2O3 3 moles O2 and 2 moles Fe2O3

2 moles Fe2O3 3 moles O2

Writing Mole-Mole Factors

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How many moles of Fe2O3 can be produced from

6.0 moles O2?

4Fe(s) + 3O2(g) 2Fe2O3(s)

Relationship: 3 mole O2 = 2 mole Fe2O3

Write a mole-mole factor to determine the moles of Fe2O3.

6.0 mole O2 x 2 mole Fe2O3 = 4.0 moles Fe2O3

3 mole O2

Calculations with Mole Factors

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Mass Calculations for Reactions


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Moles to Grams

Suppose we want to determine the mass (g) of NH3

that can form from 2.50 moles N2.

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

The plan needed would be

moles N2 moles NH3 grams NH3

The factors needed would be:

mole factor NH3/N2 and the molar mass NH3

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Moles to Grams

The setup for the solution would be:

2.50 mole N2 x 2 moles NH3 x 17.0 g NH3

1 mole N2 1 mole NH3

given mole-mole factor molar mass

= 85.0 g NH3

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The reaction between H2 and O2 produces 13.1 g water.

How many grams of O2 reacted?

2H2(g) + O2(g) 2H2O(g) ? g 13.1 g

The plan and factors would be

g H2O mole H2O mole O2 g O2

molar mole-mole molar

mass H2O factor mass O2

Calculating the Mass of a Reactant

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The setup would be:

13.1 g H2O x 1 mole H2O x 1 mole O2 x 32.0 g O2

18.0 g H2O 2 moles H2O 1 mole O2


mass H2O factor mass O2

= 11.6 g O2

Calculating the Mass of a Reactant

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Calculating the Mass of Product

When 18.6 g ethane gas C2H6 burns in oxygen, how

many grams of CO2 are produced?

2C2H6(g) + 7O2(g) 4CO2(g) + 6H2O(g) 18.6 g ? g

The plan and factors would be

g C2H6 mole C2H6 mole CO2 g CO2

molar mole-mole molar mass C2H6 factor mass CO2

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Calculating the Mass of Product

The setup would be

18.6 g C2H6 x 1 mole C2H6 x 4 moles CO2 x 44.0 g CO2

30.1 g C2H6 2 moles C2H6 1 mole CO2


mass C2H6 factor mass CO2

= 54.4 g CO2

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Active learning exercise

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Percent Yield and Limiting Reactants


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Theoretical, Actual, and Percent Yield

Theoretical yield The maximum amount of product, which is calculated

using the balanced equation.

Actual yield The amount of product obtained when the reaction takes

place.

Percent yield The ratio of actual yield to theoretical yield.

percent yield = actual yield (g) x 100 theoretical yield (g)

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You prepared cookie dough to make 5 dozen cookies.The phone rings and you answer. While you talk, a sheetof 12 cookies burn and you throw them out. The restof the cookies are okay. What is the percent yield ofedible cookies?

Theoretical yield 60 cookies possible

Actual yield 48 cookies to eat

Percent yield 48 cookies x 100 = 80% yield 60 cookies

Calculating Percent Yield

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Limiting Reactant

A limiting reactant in a chemical reaction is the

substance that

Is used up first.

Limits the amount of product that can form and stops the reaction.

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Reacting Amounts

In a table setting, there is 1plate,

1 fork, 1 knife, and 1 spoon.

How many table settings are

possible from 5 plates, 6 forks,

4 spoons, and 7 knives?

What is the limiting item?Copyright © 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings

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Reacting Amounts

Only 4 place settings are possible.

Initially Used Left overPlates 5 4 1Forks 6 4 2Spoons 4 4 0Knives 7 4 3

The limiting item is the spoon.


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Example of An Everyday Limiting ReactantHow many peanut butter sandwiches could be made from 8 slices of bread and 1 jar of peanut butter?

With 8 slices of bread, only 4 sandwiches could be made. The bread is the limiting item.


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Example of An Everyday Limiting Reactant

How many peanut butter sandwiches could be made from 8 slices bread and 1 tablespoon of peanut butter?

With 1 tablespoon of peanut butter, only 1 sandwich could be made. The peanut butter is the limiting item.


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Limiting Reactant

When 4.00 moles H2 is mixed with 2.00 moles Cl2,how many moles of HCl can form?

H2(g) + Cl(g) 2HCl (g)

4.00 moles 2.00 moles ??? moles

Calculate the moles of product that each reactant, H2 and Cl2, could produce.

The limiting reactant is the one that produces the smallest amount of product.

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Limiting Reactant

HCl from H2

4.00 moles H2 x 2 moles HCl = 8.00 moles HCl

1 moles H2 (not possible)

HCl from Cl2

2.00 moles Cl2 x 2 moles HCl = 4.00 moles HCl

1 mole Cl2 (smaller number of moles, Cl2 will be used up first)

The limiting reactant is Cl2 because it is used up first. Thus Cl2 produces the smaller number of moles of HCl.

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Check Calculations

Initially H2

4.00 moles

Cl2

2.00 moles

2HCl

0 mole

Reacted/

Formed

-2.00 moles

-2.00 moles

+4.00 moles

Left after reaction

2.00 moles

Excess

0 mole

Limiting

4.00 moles

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Limiting Reactants Using Mass

If 4.80 moles Ca are mixed with 2.00 moles N2, which is

The limiting reactant? 3Ca(s) + N2(g) Ca3N2(s)

moles of Ca3N2 from Ca

4.80 moles Ca x 1 mole Ca3N2 = 1.60 moles Ca3N2 3 moles Ca (Ca is used up)

moles of Ca3N2 from N2

2.00 moles N2 x 1 mole Ca3N2 = 2.00 moles Ca3N2

1 mole N2 (not possible)

Ca is used up when 1.60 mole Ca3N2 forms. Thus, Ca is

the limiting reactant.

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Calculate the mass of water produced when 8.00 g H2

and 24.0 g O2 react?

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

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Moles H2O from H2:

8.00 g H2 x 1 mole H2 x 2 moles H2O = 3.97 moles H2O 2.02 g H2 2 moles H2 (not possible)

Moles H2O from O2:

24.0 g O2 x 1 mole O2 x 2 moles H2O = 1.50 moles H2O 32.0 g O2 1 mole O2 O2 is limiting

The maximum amount of product is 1.50 moles H2O,which is converted to grams.

1.50 moles H2O x 18.0 g H2O = 27.0 g H2O

1 mole H2O

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1 Lecture 4 Chemical Reactions and Quantities Chemical Reactions and Equations Copyright © 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings