3 ways of representing the reaction of H 2 with O 2 to form H 2 O

3 ways of representing the reaction of H2 with O2 to form H2O

Chemical Equationsreactants products

Balancing Chemical Equations

1. Write the correct formula(s) for the reactants on the left side and the correct formula(s) for the product(s) on the right side of the equation.

Ethane reacts with oxygen to form carbon dioxide and water

C2H6 + O2 CO2 + H2O

2. Change the numbers in front of the formulas (coefficients) to make the number of atoms of each element the same on both sides of the equation. Do not change the subscripts.

2C2H6 NOT C4H12


3. Start by balancing those elements that appear in only one reactant and one product.

C2H6 + O2 CO2 + H2O start with C or H but not O

2 carbonon left

1 carbonon right multiply CO2 by 2

C2H6 + O2 2CO2 + H2O

6 hydrogenon left

2 hydrogenon right multiply H2O by 3

C2H6 + O2 2CO2 + 3H2O


4. Balance those elements that appear in two or more reactants or products.

2 oxygenon left

4 oxygen(2x2)

C2H6 + O2 2CO2 + 3H2O

+ 3 oxygen(3x1)

multiply O2 by 72

= 7 oxygenon right

C2H6 + O2 2CO2 + 3H2O72

remove fractionmultiply both sides by 2

2C2H6 + 7O2 4CO2 + 6H2O


5. Check to make sure that you have the same number of each type of atom on both sides of the equation.

2C2H6 + 7O2 4CO2 + 6H2O

Reactants Products

4 C

12 H

14 O

4 C

12 H

14 O

4 C (2 x 2) 4 C

12 H (2 x 6) 12 H (6 x 2)14 O (7 x 2) 14 O (4 x 2 + 6)

How to “Read” Chemical Equations

2 Mg + O2 2 MgO

2 atoms Mg + 1 molecule O2 makes 2 formula units MgO

2 moles Mg + 1 mole O2 makes 2 moles MgO

48.6 grams Mg + 32.0 grams O2 makes 80.6 g MgO

IS NOT

2 grams Mg + 1 gram O2 makes 2 g MgO

Chemical Equation Calculations

Reactants ProductsMolecules

Moles

MassMolecularWeight g/mol

Atoms (Molecules)

Avogadro’sNumber

6.02 x 1023

Stoichiometry (more working with ratios)

Ratios are found within a chemical equation.

2HCl + Ba(OH)2 2H2O + BaCl2 1 1

2 moles of HCl react with 1 mole of Ba(OH)2 to form 2 moles of H2O and 1 mole of BaCl2

coefficients give MOLAR RATIOS

When N2O5 is heated, it decomposes:2N2O5(g) 4NO2(g) + O2(g)

a. How many moles of NO2 can be produced from 4.3 moles of N2O5?

= moles NO2

4.3 mol N2O5

52

2

ON mol2

NO mol48.6

b. How many moles of O2 can be produced from 4.3 moles of N2O5?

= mole O2

4.3 mol N2O5

52

2

ON 2mol

O mol12.2

2N2O5(g) 4NO2(g) + O2(g)

4.3 mol ? mol

2N2O5(g) 4NO2(g) + O2(g)

4.3 mol ? mol

Mole – Mole Conversions

Units match

12

When N2O5 is heated, it decomposes:2N2O5(g) 4NO2(g) + O2(g)

a. How many moles of N2O5 were used if 210g of NO2 were produced?

= moles N2O5

210 g NO2

2

52

NO mol4

ON mol22.28

b. How many grams of N2O5 are needed to produce 75.0 grams of O2?

= grams N2O5

75.0 g O2

2

52

O 1mol

ON mol2506

2

2

NO g0.46

NO mol

2

2

O g 32.0

O mol

52

52

ON mol

ON g108

gram ↔ mole and gram ↔ gram conversions

2N2O5(g) 4NO2(g) + O2(g)210g? moles

2N2O5(g) 4NO2(g) + O2(g)75.0 g? grams

Units match

Aluminum is an active metal that when placed in hydrochloric acid produces hydrogen gas and aluminum chloride. How many grams of aluminum chloride can be produced when 3.45 grams of aluminum are reacted with an excess of hydrochloric acid?

First write a balanced equation.

Al(s) + HCl(aq) AlCl3(aq) + H2(g)2 6 2 3

Gram to Gram Conversions


Al(s) + HCl(aq) AlCl3(aq) + H2(g)2 6 2 3

Now let’s get organized. Write the information below the substances.

3.45 g ? grams

Gram to Gram Conversions


Al(s) + HCl(aq) AlCl3(aq) + H2(g)2 6 2 33.45 g ? grams

Let’s work the problem.

= g AlCl33.45 g Al

Alg 27.0

Almol

We must always convert to moles.Now use the molar ratio.

Almol 2

AlClmol 2 3

Now use the molar mass to convert to grams.

3

3

AlClmol

AlClg 133.317.0

Units match

gram to gram conversions

Methanol burns in air according to the equation

2CH3OH + 3O2 2CO2 + 4H2O

If 209 g of methanol are used up in the combustion, what mass of water is produced?

grams CH3OH moles CH3OH moles H2O grams H2O

molar massCH3OH

coefficientschemical equation

molar massH2O

209 g CH3OH1 mol CH3OH

32.0 g CH3OHx

4 mol H2O

2 mol CH3OHx

18.0 g H2O

1 mol H2Ox =

235 g H2O

3.8

Problem: Calculating Reactants and Products in a Chemical Reaction

Problem: Given the following chemical reaction between aluminum sulfide and water, if we are given 65.80 g of Al2S3: a) How many moles of water are required for the reaction? b) What mass of H2S & Al(OH)3 would be formed?

Al2S3 (s) + 6 H2O (l) 2 Al(OH)3 (s) + 3 H2S (g)

Plan: Calculate moles of aluminum sulfide using its molar mass, then from the equation, calculate the moles of water, and then the moles of hydrogen sulfide, and finally the mass of hydrogen sulfide using it’s molecular weight.

Solution: a) molar mass of aluminum sulfide =

moles Al2S3 =

Calculating Reactants and Products in a Chemical Reaction

a) cont. H2O: 0.4382 moles Al2S3 x

b)H2S: 0.4382 moles Al2S3 x

___ moles H2O 1 mole Al2S3

___ moles H2S 1 mole Al2S3

molar mass of H2S =

mass H2S =

Al(OH)3: 0.4382 moles Al2S3 x molar mass of Al(OH)3 =

mass Al(OH)3 =

Balanced reaction! Definesstoichiometric ratios!

Unbalanced (i.e., non-stoichiometric) mixture!

Limited by syrup!

6 green used up

6 red left over

Limiting Reagents

3.9

1 green + 1 red 1 green-red

Limiting Reactant

In a chemical reaction where arbitrary amounts of reactants are mixed and allowed to react, the one that is used up first is the limiting reactant. A portion of the other reactants remains.

There is a systematic procedure for finding the limiting reagent based on the reactant ratio (RR) defined as the ratio of the number of moles of a reactant to its coefficient in a balanced chemical equation. The reagent with the smallest reactant ratio is the limiting reactant.

Limiting Reagent Concept

Limiting Reagent Problems

1. Balance chemical equation2. Determine limiting reagent

• Do two separate calculations for the amount of product each reactant would produce if they were the limiting reagent

• The reactant that gives the lower number is the limiting reagent.

3. The amount of product produced is the number calculated by limiting reagent

Limiting Reagent Problems (cont)

4. Determination of the amount of excess reagent left over

• Calculate the amount of excess reagent (ER) used in chemical reaction

• Subtract the ER used from original amount of ER.

Limiting/Excess/ Reactant and Theoretical Yield Problems :

Potassium superoxide, KO2, is used in rebreathing gas masks to generate oxygen.

4KO2(s) + 2H2O(l) 4KOH(s) + 3O2(g)

a. How many moles of O2 can be produced from 0.15 mol KO2 and 0.10 mol H2O? b. Determine the limiting reactant.

4KO2(s) + 2H2O(l) 4KOH(s) + 3O2(g)

First copy down the the BALANCED equation!

Now place numerical the information below the compounds.



4KO2(s) + 2H2O(l) 4KOH(s) + 3O2(g) a. How many moles of O2 can be produced from 0.15 mol KO2 and 0.10 mol H2O?b. Determine the limiting reactant.

4KO2(s) + 2H2O(l) 4KOH(s) + 3O2(g) 0.15 mol 0.10 mol ? moles

Two starting amounts? Where do we start?

Hide

one

27



4KO2(s) + 2H2O(l) 4KOH(s) + 3O2(g) a. How many moles of O2 can be produced from 0.15 mol KO2 and 0.10 mol H2O?b. Determine the limiting reactant.

4KO2(s) + 2H2O(l) 4KOH(s) + 3O2(g) 0.15 mol 0.10 mol ? molesHide

Based on:KO2

= mol O20.15 mol KO2

2

2

KO 4mol

O mol30.1125

28


4KO2(s) + 2H2O(l) 4KOH(s) + 3O2(g)

a. How many moles of O2 can be produced from 0.15 mol KO2 and 0.10 mol H2O?b. Determine the limiting reactant.


Based on:KO2 = mol O2

0.15 mol KO2

2

2

KO 4mol

O mol30.1125

Hide

Based on: H2O

= mol O20.10 mol H2O

OH 2mol

O mol3

2

2 0.150


29



4KO2(s) + 2H2O(l) 4KOH(s) + 3O2(g) a. How many moles of O2 can be produced from 0.15 mol KO2 and 0.10 mol H2O?

Determine the limiting reactant.


Based on:KO2 = mol O2

0.15 mol KO2

2

2

KO 4mol

O mol30.1125

Based on: H2O

= mol O20.10 mol H2O

OH 2mol

O mol3

2

2 0.150

What is the theoretical yield? Hint: Which is the smallest amount? The is based upon the limiting reactant?

It was limited by theamount of KO2.

H2O = excess (XS) reactant!

Do You Understand Limiting Reagents?

In one process, 124 g of Al are reacted with 601 g of Fe2O3

2Al + Fe2O3 Al2O3 + 2Fe

Calculate the mass of Al2O3 formed.

g Al mol Al mol Fe2O3 needed g Fe2O3 needed

OR

g Fe2O3 mol Fe2O3 mol Al needed g Al needed

124 g Al1 mol Al

27.0 g Alx

1 mol Fe2O3

2 mol Alx

160. g Fe2O3

1 mol Fe2O3

x = 367 g Fe2O3

Start with 124 g Al need 367 g Fe2O3

Have more Fe2O3 (601 g) so Al is limiting reagent3.9

Use limiting reagent (Al) to calculate amount of product thatcan be formed.

g Al mol Al mol Al2O3 g Al2O3

124 g Al1 mol Al

27.0 g Alx

1 mol Al2O3

2 mol Alx

102. g Al2O3

1 mol Al2O3

x = 234 g Al2O3

2Al + Fe2O3 Al2O3 + 2Fe

3.9

Chemical Reactions in Practice: Theoretical, Actual, and Percent Yields

Theoretical yield: The amount of product indicated by the stoichiometrically equivalent molar ratio in the balanced equation. Side Reactions: These form different products that take away from the theoretical yield of the main product.

Actual yield: The amount of product that is actually obtained.

Percent yield (%Yield):

% Yield = x 100% Actual Yield (mass or moles)

Theoretical Yield (mass or moles)

or Actual yield = Theoretical yield x (% Yield / 100%)

Problem: Percent Yield

Problem: The chemical reaction between iron and water to form the iron oxide, Fe3O4 and hydrogen gas is given below. If 4.55 g of iron is reacted with sufficient water to react all of the iron to form rust, what is the percent yield if only 6.02 g of the oxide are formed?Plan: Calculate the theoretical yield and use it to calculate the percent yield, using the actual yield.Solution:

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

Moles Fe =

Theoretical moles Fe3O4 =

Theoretical mass Fe3O4 =

Percent Yield = x 100% = Actual Yield

Theoretical Yield

Problem: Percent Yield / Limiting Reactant

Problem: Ammonia is produced by the Haber process using nitrogen and hydrogen Gas. If 85.90g of nitrogen are reacted with 21.66 g hydrogen and the reaction yielded 98.67 g of ammonia what was the percent yield of the reaction.

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

Plan: Since we are given the masses of both reactants, this is a limiting reactant problem. First determine which is the limiting reactant then calculate the theoretical yield, and then the percent yield.

Solution:

moles N2 =

moles H2 =

Problem: Percent Yield / Limiting Reactant

Solution Cont. N2 (g) + 3 H2 (g) 2 NH3 (g)

We have 3.066 moles of Nitrogen, and it is limiting, therefore the theoretical yield of ammonia is:

mol NH3 =

mass NH3 =

Percent Yield = x 100%

Percent Yield = x 100% =

Actual YieldTheoretical Yield

98.67 g NH3

g NH3

Answers to Problems in the Lecture

1. (a) 2.629 moles H2O; (b) 44.81g H2S, 68.36g Al(OH)3

2. 77.61 g Ca3(PO4)2

3. 0.183 mol of AlCl3

4. 30.0 g NO

5. 95.6 %

6. 94.49 %

37

4KO2(s) + 2H2O(l) 4KOH(s) + 3O2(g) If a reaction vessel contains 120.0 g of KO2 and 47.0 g of H2O, how many grams of O2 can be produced?

4KO2(s) + 2H2O(l) 4KOH(s) + 3O2(g) 120.0 g 47.0 g ? gHide one

Based on:KO2

= g O2 120.0 g KO2

g1.71

mol2

2

KO 4mol

O mol3

2

2

O mol

O g0.32 40.51

Limiting/Excess Reactant Problem with % Yield

38

4KO2(s) + 2H2O(l) 4KOH(s) + 3O2(g) If a reaction vessel contains 120.0 g of KO2 and 47.0 g of H2O, how many grams of O2 can be produced?

4KO2(s) + 2H2O(l) 4KOH(s) + 3O2(g) 120.0 g 47.0 g ? g

Based on:KO2

= g O2 120.0 g KO2

g1.71

mol2

2

KO 4mol

O mol3

2

2

O mol

O g0.32 40.51

Based on:H2O

= g O2

Question if only 35.2 g of O2 were recovered, what was the percent yield?

yield 86.9% 100x 51.40

2.35 100x

ltheoretica

actual

Hide

47.0 g H2O

OH g 02.18

OH mol

2

2

OH mol 2

O mol 3

2

2

2

2

O mol

O g0.32 125.3

Limiting/Excess Reactant Problem with % Yield

39

If a reaction vessel contains 120.0 g of KO2 and 47.0 g of H2O, how many grams of O2 can be produced?

4KO2(s) + 2H2O(l) 4KOH(s) + 3O2(g) 120.0 g 47.0 g ? g

Based on:KO2

= g O2 120.0 g KO2

g1.71

mol2

2

KO 4mol

O mol3

2

2

O mol

O g0.32 40.51

Based on:H2O

= g O247.0 g H2O

OH g 02.18

OH mol

2

2

OH mol 2

O mol 3

2

2

2

2

O mol

O g0.32 125.3

Determine how many grams of Water were left over.The Difference between the above amounts is directly RELATED to the XS H2O.

125.3 - 40.51 = 84.79 g of O2 that could have been formed from the XS water.

= g XS H2O84.79 g O2

2

2

O g 32.0

O mol

2

2

O mol 3

OH mol 2

OH mol 1

OH g 02.18

2

2 31.83

The decomposition of KClO3 is commonly used to prepare small amounts of O2 in the laboratory:

How many grams of O2 can be prepared from 4.50 g of KClO3?

Propane, C3H8, is a common fuel used for cooking and home heating. What mass of O2 is consumed in the combustion of 1.00 g of propane?

How many moles of NH3 can be formed from 3.0 mol of N2 and 6.0 mol of H2?

The most important commercial process for converting N2 from the air into nitrogen-containing compounds is based on the reaction of N2 and H2 to form ammonia (NH3):

Consider the reaction

A mixture of 1.50 mol of Al and 3.00 mol of Cl2 is allowed to react. (a) Which is the limiting reactant? (b) How many moles of AlCl3 are formed? (c) How many moles of the excess reactant remain at the end of the reaction?

A strip of zinc metal having a mass of 2.00 g is placed in an aqueous solution containing 2.50 g of silver nitrate, causing the following reaction to occur:

(a) Which reactant is limiting? (b) How many grams of Ag will form? (c) How many grams of Zn(NO3)2 will form? (d) How many grams of the excess reactant will be left at the end of the reaction?

Molarity

Molarity is a term used to express concentration. The units of molarity are moles per liter (it is abbreviated as a capital M).

When working problems, it is a good idea to change M into its units.

mL 1000

moles

Liter

moles M

A solution is prepared by dissolving 3.73 grams of AlCl3 in water to form 200.0 mL solution. A 10.0 mL portion of the solution is then used to prepare 100.0 mL of solution. Determine the molarity of the final solution.

What type of problem(s) is this?

Molarity followed by dilution.

Solutions

A solution is prepared by dissolving 3.73 grams of AlCl3 in water to form 200.0 mL solution. A 10.0 mL portion of the solution is then used to prepare 100.0 mL of solution. Determine the molarity of the final solution.

1st:= mol L

3.73 g

g 133.4

mol200.0 x 10-3 L

0.140

2nd: M1V1 = M2V2

(0.140 M)(10.0 mL) = (? M)(100.0 mL)0.0140 M = M2

molar mass of AlCl3

dilution formula

final concentration

Solutions

50.0 mL of 6.0 M H2SO4 (battery acid) were spilled and solid NaHCO3 (baking soda) is to be used to neutralize the acid. How many grams of NaHCO3 must be used?

H2SO4(aq) + 2NaHCO3 2H2O(l) + Na2SO4(aq) + 2CO2(g)

Solution Stoichiometry

50.0 mL6.0 M

L

mol 6.0

? g

Look! A conversion factor!




=

Our Goal

50.0 mL6.0 M

L

mol 6.0

? g




=

Our Goal

= g NaHCO3

H2SO4

50.0 mL

1000mL

SOH mol 6.0

42SOH

42

1 molH2SO4

NaHCO3

2 molNaHCO3

84.0 gmolNaHCO3

50.4

Solution Stoichiometry:

Determine how many mL of 0.102 M NaOH solution are needed to neutralize 35.0 mL of 0.125 M H2SO4 solution.

First write a balancedEquation.

____NaOH + ____H2SO4 ____H2O + ____Na2SO4 2 1 2 1


Determine how many mL of 0.102 M NaOH solution is needed to neutralize 35.0 mL of 0.125 M H2SO4 solution.

Now, let’s get organized. Place numerical Information and accompanying UNITS below each compound.

____NaOH + ____H2SO4 ____H2O + ____Na2SO4 2 1 2 10.102 M

L

mol

? mL

35.0 mL

mL 1000

mol 0.125

L

mol 0.125

Since 1 L = 1000 mL, we can use this to save on the number of conversions

Our Goal

Determine how many mL of 0.102 M NaOH solution is needed to neutralize 35.0 mL of 0.125 M H2SO4 solution.

Now let’s get to work converting.

____NaOH + ____H2SO4 ____H2O + ____Na2SO4 2 1 2 1

0.102 ML

mol

? mL

35.0 mL

mL1000

mol 0.125

L

mol 0.125

= mL NaOH

H2SO4

35.0 mL H2SO4

0.125 mol 1000 mL H2SO4

NaOH2 mol1 mol H2SO4

1000 mL NaOH0.102 mol NaOH

85.8

Units Match


shortcut

54

What volume of 0.40 M HCl solution is needed to completely neutralize 47.1 mL of 0.75 M Ba(OH)2?

1st write out a balanced chemicalequation


55

What volume of 0.40 M HCl solution is needed to completely neutralize 47.1 mL of 0.75 M Ba(OH)2?

2HCl(aq) + Ba(OH)2(aq) 2H2O(l) + BaCl2

0.40 M 47.1 mL0.75 M? mL

= mL HCl

Ba(OH)2

47.1 mL

2

2

Ba(OH)

Ba(OH)

mL 1000

0.75mol

1 mol Ba(OH)2

HCl2 mol

0.40 mol HCl

HCl1000 mL 176

Units match


3 ways of representing the reaction of H 2 with O 2 to form H 2 O

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Transcript of 3 ways of representing the reaction of H 2 with O 2 to form H 2 O