1 stoichiometry 2 general approach for problem solving 1. clearly identify the goal or goals and the...

1

STOICHIOMETRYSTOICHIOMETRY

2

General Approach For Problem Solving

1. Clearly identify the Goal or Goals and the UNITS involved. (starting and ending unit)

2. Determine what is given and the UNITS.

3. Use conversion factors (which are really ratios) and their UNITS to CONVERT what is given into what is desired.

3

Sample problem for general problem solving.

Sam has entered into a 10 mile marathon. Use ALL of the following conversions (ratios) to determine how many inches there are in the race.5280 ft = 1 mile; 12 inches = 1 ft

1. What is the goal and what units are needed?

Goal = ______ inches

2. What is given and its units?

10 miles

3. Convert using factors (ratios).

10 miles = inches

mile 1

ft 5280ft 1

inches 12 633600

Units match

Given GoalConvert

Menu

4

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

Information given by chemical Information given by chemical equationsequations

2 C6H6 (l) + 15 O2 (g) 12 CO2 (g) + 6 H2O (g)

In this equation there are In this equation there are 22 molecules of benzene reacting with molecules of benzene reacting with 1515 molecules of oxygen to produce molecules of oxygen to produce 1212 molecules of carbon dioxide and molecules of carbon dioxide and 66

molecules of watermolecules of water.

This equation could also be read as This equation could also be read as 22 moles of benzene reacts with moles of benzene reacts with 1515 moles of moles of oxygen to produce oxygen to produce 1212 moles of carbon dioxide and moles of carbon dioxide and 66 moles of water. moles of water.

Since the relationship between the actual number of molecules and the number of moles present is 6.02 x 1023, a common factor between all species

involved in the equation, a MOLE RATIOMOLE RATIO relationship can be discussed.

Information given by chemical Information given by chemical equationsequations

2 C6H6 (l) + 15 O2 (g) 12 CO2 (g) + 6 H2O (g)

The The MOLE RATIOMOLE RATIO for benzene and oxygen is 2 : 15. It can be written as: for benzene and oxygen is 2 : 15. It can be written as: 2 moles C6H6 or as 15 moles of O2

15 moles O2 2 moles of C6H6

The The MOLE RATIOMOLE RATIO for oxygen and carbon dioxide is 15 : 12. for oxygen and carbon dioxide is 15 : 12. It can be written as:It can be written as:

12 moles CO2 or as 15 moles of O2

15 moles O2 12 moles of CO2

NOTE: The MOLE RATIO is used for converting moles of one substance into moles of another substance. Without the balanced equation there is no

other relationship between two different compounds.

Using the mole ratio to relate the moles of one compound to the moles of another compound is the part of chemistry called

STOICHIOMETRY !!!!!STOICHIOMETRY !!!!!

2 H2 H22 (g)(g) ++ O O22 (g)(g) 2 H2 H22O O (g)(g)

Q. How many mole of hydrogen are necessary to react with 2 moles of oxygen in order to produce exactly 4 moles of water?

A. 2 mol O2 (2 moles H2 / 1 mole O2) = 4 mole H2


The Stoichiometry Flow The Stoichiometry Flow ChartChart

M a s s o f A m ole o f A m o les o f B M a s s o f B

Use Molar mass (A)

Use mole ratio from equation

Use Molar mass (B)


2 H2 H22 (g)(g) + O + O22 (g)(g) 2 H 2 H22O O (g)(g)

Q1. How many moles of hydrogen are necessary to react with 15.0 g of oxygen?

A. 15.0g O2 (1 mole O2

) ( 2 mole H2) = 0.938 moles H2

32.0 g 1 mole O2

Q2. How many grams of hydrogen are necessary to react with 15.0 g of oxygen?

A. 15.0g O2 (1 mole O2

) ( 2 mole H2) ( 2.016 g H

2) = 1.89 g H2 32.0 g 1 mole O

2 1 mole H2


2 H2 H22 (g)(g) + O + O22 (g)(g) 2 H 2 H22O O (g)(g)Q3. How many grams of water are produced from 15.0 g of oxygen?

A. 15.0g O2 (1 mole O2

) ( 2 mole H2

O) ( 18.0 g H2

O) =16.9 g H2O 32.0 g 1 mole O

2 1 mole H2

O

Q4. How much hydrogen and oxygen is needed to produce 25.0 grams of water?

A. 25.0g H2O (1 mole H2

O ) ( 2 mole H2) ( 2.016 g H

2) = 2.80 g H2 18.0 g 2 mole H

2O 1 mole H

2

A. 25.0g H2O (1 mole H2

O ) ( 1 mole O2) ( 32.0 g O

2) = 22.2 g O2 18.0 g 2 mole H

2O 1 mole O

2

Notice that the Law of Conservation of Mass still applies.

How many grams of solid are formed How many grams of solid are formed when 10.0 g of lead reacts with excess when 10.0 g of lead reacts with excess

phosphoric acid?phosphoric acid?1. Write the chemical equation: Pb + H3PO4 ?

You recognize that this is a single displacement (replacement) reaction. So Pb (a metal) will displace (replace) H (the cation).

Pb + H3PO4 Pb3(PO4)2 + H2

2. Balance the equation: 3 Pb+2 H3PO4 Pb3(PO4)2 + 3 H2

3. Make a list under the appropriate substance

3 Pb+2 H3PO4 Pb3(PO4)2 (s) + 3 H2 (g)

10.0g m=?

Start with what is given:

10.0gPb (1 mole Pb)( 1 mole Pb3

(PO4

)2)(811 g Pb

3(PO

4)2) = 13.1 g Pb13.1 g Pb33(PO(PO44))22

207 g Pb 3 mole Pb 1 mole Pb3

(PO4

)2

PRACTICE PROBLEM # 181. How many grams of gas can be produced from 0.8876 moles of

HgO?

2 HgO 2 Hg + O2

2. How many moles of fluorine are required to produce 12.0 grams of KrF6? Given the equation: Kr + 3 F2 KrF6

3. How many grams of Na2CO3 will be produced from the thermal decomposition of 250.0 g of NaHCO3?

4. How many grams of CO2 can be produced by the reaction of 75.0 grams of C2H2 with excess oxygen?

5. Cu + 2 AgNO3 Cu(NO3)2 + 2 Ag.

How many grams of silver is produced when 125.0 g of copper is reacted with excess silver nitrate solution?

14.20g

0.182 mol

157.7 g

254 g

424.9 g

GROUP STUDY PROBLEM # 18______1. How many grams of liquid product can be produced from 3.55

moles of HgO?

2 HgO 2 Hg + O2

______2. How many moles of fluorine are required to produce 3.0 grams of KrF6? Given the equation: Kr + 3 F2 KrF6

______3. How many grams of Na2CO3 will be produced from the decomposition of 20.0g of NaHCO3?

______4. How many grams of O2 are needed to combust 55.0 grams of C2H4?

______5. Cu + 2 AgNO3 Cu(NO3)2 + 2 Ag.

How many grams of silver is produced when 50.0 g of copper is reacted with excess silver nitrate solution?

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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

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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

16

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

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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

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

Let’s work the problem.

= g AlCl3

3.45 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

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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

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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

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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

mol

200.0 x 10-3 L0.140

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

22

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

23

50.0 mL

6.0 M

L

mol 6.0

? g

Look! A conversion factor!




=

Our Goal

24

50.0 mL

6.0 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

25

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

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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 1

0.102 ML

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

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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

28

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 chemical

equation


29

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


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Solution Stochiometry Problem:

A chemist performed a titration to standardize a barium hydroxide solution. If it took 23.28 mL of 0.135 M hydrochloric acid to neutralize 25.00 mL of the barium hydroxide solution, what was the concentration of the barium hydroxide solution in moles per liter (M)?

First write a balanced chemical reaction.

____HCl(aq) + ____Ba(OH)2(aq) ____H2O(l) + ____BaCl2(aq)2 1 2 1

23.28 mL

0.135 mol L

25.00 mL

? mol L

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A chemist performed a titration to standardize a barium hydroxide solution. If it took 23.28 mL of 0.135 M hydrochloric acid to neutralize 25.00 mL of the barium hydroxide solution, what was the concentration of the barium hydroxide solution in moles per liter (M)?

____HCl(aq) + ____Ba(OH)2(aq) ____H2O(l) + ____BaCl2(aq)2 1 2 1

23.28 mL

0.135 mol L

25.00 mL

? mol L

= mol Ba(OH)2

L Ba(OH)2

25.00 x 10-3 L Ba(OH)2

Units Already Match on Bottom!

HClmL 23.28

HCl

HCl

mL 1000

mol 0.135

HCl

Ba(OH)

mol 2

mol l2 0.0629

Units match on top!

32

48.0 mL of Ca(OH)2 solution was titrated with 19.2 mL of 0.385 M HNO3. Determine the molarity of the Ca(OH)2 solution.

We must first write a balanced equation.


33

48.0 mL of Ca(OH)2 solution was titrated with 19.2 mL of 0.385 M HNO3. Determine the molarity of the Ca(OH)2 solution.

Ca(OH)2(aq) + HNO3(aq) H2O(l) + Ca(NO3)2(aq)2 248.0 mL 19.2 mL

0.385 ML

mol 0.385

= mol(Ca(OH)2)

L (Ca(OH)2)

19.2 mLHNO3

3

3

HNO

HNO

mL 1000

mol0.385

3

2

HNO 2mol

Ca(OH) 1mol

48.0 x 10-3L

? M

units match!

0.0770


34

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.

35



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

36



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 O2

0.15 mol KO2

2

2

KO 4mol

O mol30.1125

37


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 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


38



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.

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



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!

39

Theoretical yield vs. Actual yield

Suppose the theoretical yield for an experiment was calculated to be 19.5 grams, and the experiment was performed, but only 12.3 grams of product were recovered. Determine the % yield.

Theoretical yield = 19.5 g based on limiting reactant

Actual yield = 12.3 g experimentally recovered

100x yield ltheoretica

yield actual yield %

yield 63.1% 100x 19.5

12.3 yield %

40

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 g0.3240.51

Limiting/Excess Reactant Problem with % Yield

41

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


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 g0.3240.51

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 g0.32125.3

Limiting/Excess Reactant Problem with % Yield

42


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 g0.3240.51

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 g0.32125.3

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

43

Calculate the molarity of a solution prepared by dissolving 25.6 grams of Al(NO3)3 in 455 mL of solution.

L

mol 0.264

L 10x 455g 213

moleg 25.63-

After you have worked the

problem, click here to see

setup answer

Try this problem (then check your answer):

1 stoichiometry 2 general approach for problem solving 1. clearly identify the goal or goals and the...

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