STOICHIOMETRY Chapter 12 12.pdfthe mole ratio and molar mass 2. Example: In the reaction 2Na + Cl 2 → 2NaCl, find the number of grams of sodium chloride produced when 1.25 mol of

Chapter 12 STOICHIOMETRY

A. Stoichiometry: study of

quantitative relationships among

masses and volumes of

reactants and products in a

chemical reaction

1. Used to make predictions about:

a. How much product is obtained from

given amount of reactant

b. How much reactant is needed to give

required amount of product

c. How much of one reactant is required

to give optimum result with given

amount of another reactant

I. STOICHIOMETRY BACKGROUND

B. Mole-Mass Relationships in Reactions

1. Chemical equations can be used to find molar ratios & mass ratios

2. Coefficients represent the number of moles of a specific type of particle in a chemical reaction

3. Example: 4Fe + 3O2 → 2Fe2O3

a. Shows that four moles of iron react with three moles of oxygen to produce two moles of iron (III) oxide

b. Molar masses can be used with the coefficients to illustrate the law of conservation of mass = 223.4 g Fe = 96.00 g O2

= 319.4 g Fe2O3

c. 223.4 g Fe + 96.00 g O2 = 319.4 g Fe2O3

d. 319.4 g = 319.4 g ✓

4 mol Fe 55.85 g Fe

1 mol

2 mol Fe2O3 159.7 g Fe2O3

1 mol Fe2O3

3 mol O2 32.00 g O2

1 mol O2

4. Mole ratio: ratio between the numbers of moles of any two

substances in a chemical reaction

a. Writing & identifying mole ratios is critical in calculations based upon

chemical equations

b. To write mole ratios:

i. Identify each term in the equation

ii. Write all possible reactant / reactant, reactant / product, product / product ratios

c. Example: 2KClO3 → 2KCl + 3O2

i. Terms: 2KClO3, 2KCl, 3O2

ii. Ratios:

2 mol KClO3

2 mol KCl

2 mol KClO3

3 mol O2

2 mol KCl

2 mol KClO3

3 mol O2

2 mol KClO3

2 mol KCl

3 mol O2

3 mol O2

2 mol KCl

p. 356 #1

p. 357 #2-8

Yes, there are a few questions when you have to write

equations from words -- you learned how to do this, don’t freak

out, look at Chapter 10 notes

HOMEWORK

A. Mole-to-Mole Conversions

1. The number of moles of one substance in a chemical reaction can

be determined from a known number of moles & a mole ratio

2. Example: In the reaction 2K + 2H2O → 2KOH + H2, 0.400 mol of

potassium are used. How many moles of water are needed? How

many moles of each product will be produced?

moles H2O = 0.400 mol

II. STOICHIOMETRIC CALCULATIONS

moles of unknown =

moles of known mole ratio: unknown

mole ratio: known

moles H2O =

0.400 mol K 2 mol H2O

2 mol K

moles KOH = 0.400 mol

moles H2 = 0.200 mol

moles KOH =

0.400 mol K 2 mol KOH

2 mol K

moles H2 =

0.400 mol K 1 mol H2

2 mol K

B. Mole-to-Mass Conversions

1. Moles of substance can be converted to mass of substance using

the mole ratio and molar mass

2. Example: In the reaction 2Na + Cl2 → 2NaCl, find the number of

grams of sodium chloride produced when 1.25 mol of chlorine gas is

reacted.

146.1 g NaCl

mass unknown =

moles of known mole ratio: unknown molar mass

unknown

mole ratio: known 1 mol unknown

mass unknown =

1.25 mol Cl2 2 mol NaCl 58.44 g NaCl

1 mol Cl2 1 mol NaCl

C. Mass-to-Mole-to-Mass Conversions

1. T-Chart setup

a. Write a balanced chemical equation.

b. In T-chart, put mass of given in upper left

c. In second column, convert to moles of the given substance using molar

mass of the substance given

d. In third column, convert to the moles of unknown from the moles of the

given substance using molar ratio from coefficients in balanced equation

e. In last column, determine the mass of the unknown substance using

molar mass of the unknown

mass

unknown =

grams of

known (given) 1 mol known

mole ratio:

unknown

molar mass

unknown

molar mass

known

mole ratio:

known 1 mol unknown

2. Example: ammonium nitrate decomposes to produce nitrogen oxide

gas and water. Find the mass of water produced from the

decomposition of 25.00 g ammonium nitrate.

a. Equation: NH4NO3 → N2O + 2H2O

mass H2O = 11.26 g

CLASSWORK: p. 362 #13-14

mass H2O =

25.00 g NH4NO3 1 mol NH4NO3 2 mol H2O 18.02 g H2O

80.04 g

NH4NO3 1 mol NH4NO3 1 mol H2O

how many sandwiches can be made?

III. LIMITING REACTANTS

A. In actual chemical reactions, it is rare that reactants are

present in the exact ratios specified in the balanced

equation

1. Usually one of the reactants will not be present in large enough

quantity to react with the other reactant(s), and the reaction will “run

out” of this reactant

a. This is called the limiting reactant because it limits the amount of

product that can be formed

b. The reaction stops when all of the limiting reactant is gone

2. The reactant that is “left over” -- that is, the reactant that remains

after all of the limiting reactant is used -- is the excess reactant

B. To determine how

much product will be

obtained in a reaction,

we need to determine

which reactant is

l imited & how much

product will result

from a cer tain amount

of the limiting reactant

Use

molar

mass

Use

molar

mass

Use

molar

mass

C. Steps to calculations when the reactant is limited:

1. Write a balanced equation

2. Convert given masses of reactants to moles

3. Use moles of reactants & balanced mole ratio to determine how

many moles of product will be produced from each reactant

4. The reactant that produces the smallest number of moles of product

is the limiting reactant

5. To determine amount of product:

6. To determine how much of the excess reactant was used:

moles of limiting reactant mole ratio: product molar mass product

mole ratio: LR 1 mol product

moles of limiting reactant mole ratio: ER molar mass ER

mole ratio: LR 1 mol ER

D. Examples!

1. Find the mass of Al2S3 produced when 9.00 g of aluminum reacts

with 8.00 g of sulfur.

a. Write a balanced equation

b. Convert given masses of reactants to moles

c. Use moles of reactants & balanced mole ratio to determine how many

moles of product will be produced from each reactant

d. The reactant that produces the smallest number of moles of product is

the limiting reactant

e. To determine amount of product:



2. How many grams of NH3 can be produced from the reaction of 28.00

g of N2 and 25.00 g of H2?






the limiting reactant

e. To determine amount of product:



3. Given 15.00 g of each reactant in the reaction below, what mass of

the excess reactant will be used in the completed reaction?

__Pb(NO3)2 + __HCl → __PbCl2 + __HNO3






the limiting reactant; the other reactant is in excess

e. To determine how much of the excess reactant was used:



4. Given 200.0 g of each reactant in the reaction below, what mass of

the excess reactant will remain once the reaction is complete?

__AgNO3 + __Na3PO4 → __Ag3PO4 + __NaNO3






the limiting reactant; the other reactant is in excess

e. To determine how much of the excess reactant was used:

f. To determine how much excess reactant remains: subtract the answer

above from the original mass of the excess reactant



p. 368 #20-21

p. 369 #22-26

Read Lab 11.1, do pre-lab

HOMEWORK

A. In reality, chemical reactions are not per fect -- samples are

not pure, losses happen, and conditions are not ideal. It is

rarely that you get exactly the amount calculated.

B. Percent yield: the actual amount of product obtained

expressed as a percentage of the calculated theoretical yield

of that product.

IV. PERCENT YIELD

percent yield =

actual yield

x 100% theoretical yield

1. Example 1: 35.25 g of product is produced in a reaction. The

predicted amount is 37.23 g. Find the % yield of this product.

a. We are given both the actual yield (35.2 g) and the theoretical yield (37.2

g), so we can immediately use the percent yield equation.

b. percent yield = 94.68%

percent yield =

35.25 g

x 100% 37.23 g

2. Example 2: potassium chromate reacts with silver nitrate to produce

solid silver chromate and potassium nitrate in the equation K 2CrO4 +

2AgNO3 → Ag2CrO4 + 2KNO3. Find the % yield if 5.00 g of silver

nitrate produces 3.52 g of silver chromate in a laboratory

experiment.

a. We are given the actual yield (3.52 g), but not the theoretical yield.

b. We need to find the theoretical yield using molar masses & mole ratio

c. theoretical yield = 4.88 g Ag2CrO4

d. Now that we have the theoretical yield (4.88 g), we can use that with the

actual yield (3.52 g) to determine percent yield.

e. percent yield = 72.13%

5.00 g AgNO3 1 mol AgNO3 1 mol Ag2CrO4 331.7 g Ag2CrO4

169.9 g AgNO3 2 mol AgNO3 1 mol Ag2CrO4

percent yield = 3.52 g

x 100% 4.88 g

p. 372: #27-29

Equation #28: Cu + 2AgNO3 → 2Ag + Cu(NO3)2

Equation #29: Zn + I2 → ZnI2

p. 373: #30-34

p. 377: vocab

p. 378: concept map [DRAW]

p. 378: mastering concepts #36-50

p. 378-81: mastering problems #90

HONORS ONLY p. 382: thinking critically #99-103

p. 383: standardized test practice #1-7

REVIEW

Documents

STOICHIOMETRY Chapter 12 12.pdfthe mole ratio and molar mass 2. Example: In the reaction 2Na + Cl 2 → 2NaCl, find the number of grams of sodium chloride produced when 1.25 mol of