STOICHIOMETRY:  tells relative amts of reactants & products in a chemical reaction

• Given an amount of a substance involved in a chemical reaction, we can figure out the amount of the other substances are needed or produced
• Always compare the number of MOLES
• 4 Fe  +  3 O2  à  2 Fe2O3
• If I have 4 moles of Fe, I need 3 moles of O2 in order to produce 2 moles of Fe2O3.
• Comparison of coefficients = MOLE RATIO
• In the above eqn, what is the mole ratio between Fe2O3 and O2?
• Use coefficients…                       2 : 3  or  2 to 3

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To solve stoichiometry problems…               ALWAYS!!!!!!!!!!!

** WRITE THE BALANCED EQN & GIVEN INFORMATION! **

1.)  Find moles of given element or compound.

* Use molar mass of given substance, if problem gives you grams.

2.)  Use mole ratio (coefficients) from balanced equation.

* Use molar mass of unknown substance, if question asks for grams.

So  here’s an example:
If 34.32 grams of acetylene (C2H2) are burned in air, how many moles of CO2 can be formed?

2      C2H2  +  5  O2  à  4  CO2  +  2  H2O

34.32 g                        ? moles

1.) 34.32 g C2H2 | 1 mole C2H2 = 1.32 moles C2H2              C: 2 x 12.0 = 24.0

| 26 grams                                                        H: 2 x  1.0 =   2.0 +

26.0

2.) 1.32 moles C2H2 =  x moles CO2                      2 x = 5.28
2                       4                                  x = 2.64 moles CO2

Problem asked for moles of CO2, so that is your answer.  Be sure to round for significant figures.
Answer should be reported as 2.640 moles CO2

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Example 2:

How many grams of Na2CO3 are needed to react completely with 6.21 moles of Ca(OH)2?

Na2CO3  +  Ca(OH)2  à  2  NaOH  +  CaCO3

? g         6.21 moles

1.) Step 1 is already done for you!  The number of moles is given in the problem!  YAY!

2.) 6.21 moles Ca(OH)2 = x moles Na2CO3            x = 6.21 moles Na2CO3

1                              1

3.) 6.21 moles Na2CO3 |  106 grams   =
658 g Na2CO3

|    1 mole

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Example 3:

If 27.8 grams of Na2SO4 are reacted with excess Al2(CO3)3, how many grams of Al2(SO4)3 will be formed?

3      Na2SO4  +  Al2(CO3)3  à  Al2(SO4)3  +  3  Na2CO3

27.8 g                               ?  g

1.)  27.8 g Na2SO4 | 1 mole Na2SO4 = 0.196 moles Na2SO4                  Na: 2 x 23.0 = 46.0

| 142.1 grams                                                             S:  1 x 32.1 =  32.1

O: 4 x 16.0 =  64.0 +

2.)  0.196 moles Na2SO4  = x mole Al2(SO4)3                                                                 142.1
3                                 1                               3 x = 0.196
x = 0.0653 moles Al2(SO4)3

3.)  0.0653 mole Al2(SO4)3 |  342.3 g Al2(SO4)3 =
22.4 g Al2(SO4)3                   Al: 2 x  27.0 =   54.0                                                         | 1 mole Al2(SO4)3                                                                          S:  3 x  32.1 =   96.3

O: 12 x 16.0 = 192.0 +                                                                                                                                                                     342.3

Practice Problem:

How many grams of calcium nitrate are formed when 57.9 grams of iron (III) nitrate react with excess calcium hydroxide according to the following equation?

___ Ca(OH)2  +  ___ Fe(NO3)3  à  ___ Ca(NO3)2  +  ___ Fe(OH)3