Stoichiometry Chapter 11. Stoichiometry = the study of quantitative relationships between the amounts of reactants used and products formed by a chemical.

StoichiometryChapter 11

Stoichiometry = the study of quantitative relationships between the amounts of reactants used and products formed by a chemical reaction

– Based on the Law of Conservation of Mass• Mass is neither created nor destroyed in a

chemical reaction

Law of Conservation of Mass

4Fe + 3O2 2Fe2O3

Reactants4 mol Fe x 55.85g Fe = 223.4g Fe

1 mol Fe

3 mol O2 x 32.00g O2 = 96.00g O2

1 mol O2

Total 319.4 g

Products2 mol Fe2O3 x 159.7g Fe2O3 = 319.4 g Fe2O3

1 mol Fe2O3

Total 319.4 g

Balancing Equations Review

• Change the coefficients to make the number of atoms of each element equal on both sides of the equation

*NEVER change a subscript…doing so changes the identity of the substance

• Write the coefficients in their lowest possible ratio

• Check your work

A few things you do NOT do…

H2 + O2 ---> H2O

1. You cannot change a subscriptH2 + O2 ---> H2O2

2. You cannot place a coefficient in the middle of a formula

H2 + O2 ---> H22O

3. Make sure that your final set of coefficients are all whole numbers with no common factors other than one

4 H2 + 2 O2 ---> 4 H2O

Balancing Equations Review

Balance this equation

NaOH(aq) + CaBr2(aq) Ca(OH)2(s) + NaBr(aq)

2NaOH(aq) + CaBr2(aq) Ca(OH)2(s) + 2NaBr(aq)

Review: Mole Ratio or (Conversion Factors)

Mole Ratio = ratio between the numbers of moles of ANY two substances in a balanced chemical equation

2Al + 3Br2 2AlBr3

What mole ratios can be written for this reaction?

2mol Al 3mol Br2

3molBr2 2mol AlBr3

2molAl 2molAlBr3

2molAlBr3 2molAl

3mol Br2 2molAlBr3

2mol Al 3molBr2

Helpful Hints• ALWAYS make sure the chemical

equation is balanced!!!• Moles are ALWAYS involved when solving

stoichiometric problems• The mole ratio is needed to convert from

one substance to another in a balanced chemical equation

• If you are in doubt about how to proceed in solving a problem,

GO TO MOLES FIRST!!

Mole to Mole Conversions

How many moles of hydrogen are produced when 0.0400 mole of potassium is used?

2K + 2H20 2KOH + H2

1. Identify the known… K

2. Identify the unknown…H2

3. To solve this problem, you need to know how the unknown moles of H2 are related to the

know moles of K

4. The correct ratio should have the moles of unknown in the numerator and the moles of

the known should be in the denominator

1mol H2

2mol K

5. This mole ratio can be used to convert the known number of moles of K to a number of

moles of H2

0.0400mol K x 1mol H2 = 0.0200mol H2

2mol K

If you put 0.0400mol K into water, 0.0200mol H2 will be produced

Try one…

C3H8 + 5O2 3CO2 + 4H2O

How many moles of CO2 are produced when 10.0 moles of propane are burned in

excess oxygen in a gas grill?

Mole to Mass Conversions

• A mole to mass conversion is done when you know the number of moles of a reactant/product and you want to calculate the mass of another reactant/product

Example:

Determine the mass of NaCl produced when 1.25 moles of chlorine gas reacts

vigorously with sodium.

2Na + Cl2 2NaCl

1. Unknown…NaCl

2. Known…Cl23. Mole ratio…. 2mol NaCl

1 mol Cl24. Multiply the given by the mole ratio

1.25mol Cl2 x 2mol NaCl = 2.50mol NaCl

1 mol Cl2

Example: continued…

Determine the mass of NaCl produced when 1.25 moles of chlorine gas reacts vigorously with sodium.

2Na + Cl2 2NaCl

5. Multiply the moles of NaCl by the molar mass of NaCl

2.50mol NaCl x 58.45g NaCl =

1mol NaCl

= 146g NaCl produced

Try one…

Sodium chloride is decomposed into the elements sodium and chlorine by means of electrical energy. How many grams of

chlorine gas can be obtained from 2.50mol NaCl?

2NaCl 2Na + Cl2

Mass to Mass Conversions

Example:

Determine the mass of H2O produced from the decomposition of 25.0g of solid

ammonium nitrate.

NH4NO3 N2O + 2H2O

1. Unknown…H2O

2. Known… NH4NO3

3. Convert the grams of the given to moles

25.0g NH4NO3 x 1mol NH4NO3

80.06g NH4NO3

= 0.312mol NH4NO3

Example:

Determine the mass of H2O produced from the decomposition of 25.0g of solid

ammonium nitrate.

NH4NO3 N2O + 2H2O

4. Mole ratio… 2mol H2O

1mol NH4NO3

5. Multiply by the moles of NH4NO3 previously calculated

0.312mol NH4NO3 x 2mol H2O = 0.624mol H2O

1mol NH4NO3

Example:

Determine the mass of H2O produced from the decomposition of 25.0g of solid

ammonium nitrate.

NH4NO3 N2O + 2H2O

6. Calculate the mass of H20 using the molar mass conversion factor

0.624mol H2O x 18.02g H2O = 11.2g H2O

1mol H2O

Courtesy of Chemistry Matter and Change

Glencoe McGraw-Hill

Chemistry for DummiesJohn T. Moore

www.karentimberlake.com