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

 

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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. 

3.)  Find answer. 

       * 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

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 


            Route                            Molar                                                    Molar              Route
            6.022 x 1023                   Mass Ave.                                          Mass Ave.            6.022 x 1023

 

 

 

 

 

 

 

 

 

 

PERCENT YIELD

~ compares the actual amount of product that you made (in an experiment) to the amount of product you should have made (according to calculations)

 

~ ACTUAL YIELD:  amount of product that you made in an experiment; when given in the problem, the amount (grams, atoms/molecules, moles) given will be associated with a product of the reaction

 

~ THEORETICAL YIELD:  amount of product that you should have made (according to calculations);  amount given with a reactant should be used to calculate theoretical yield

 

 

~          % YIELD =        ACTUAL YIELD     X 100

                                THEORETICAL YIELD

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EXAMPLE
What is the percent yield if a student makes 5.2 grams of carbon dioxide by decomposing 9.5 grams of aluminum bicarbonate?
                        ___ Al(HCO3)3  à  ___ Al2O3  +  ___ CO2  +  ___ H2O

 

ANSWER:
                        _2_ Al(HCO3)3  à  ___ Al2O3  +  _6_ CO2  +  _3_ H2O

                                 9.5 g                                   ? g

* I am given the actual yield in the problem;  I need to find the theoretical yield. *

9.5 g Al(HCO3)3 | 1 mole Al(HCO3)3 |     6 moles CO2     |   44 g CO2    = 6.0 g CO2
                       |  210 grams         | 2 moles Al(HCO3)3 | 1 mole CO2

So, 6.0 g is the theoretical yield.

% yield =  5.2 g  x 100 = 87 %
                6.0 g

 

PRACTICE PROBLEM

What is the percent yield if a student produces 3.7 grams of water by reacting 7.5 grams of oxygen with excess hydrogen?

___ H2  +  ___ O2  à  ___ H2O

 

 

 

Limiting Reactant

Limiting Reactant:  reactant that will be consumed (used up) first;  limits the amount of product that can be made (produced)

 

Excess Reactant:  reactant that will not be completely consumed by reacting with all of the limiting reactant;  there will be some of this reactant left over after the reaction is complete

 

How is a limiting reactant problem different from a regular stoichiometry problem?  There are amounts (grams, atoms/molecules, moles) given with both of the reactants.

 

How are the calculations different from regular stoichiometry problems?  Once you determine the limiting reactant, the calculations are exactly the same.

 

How can I determine which is the limiting reactant?  Find the number of moles of each reactant.  Divide the number of moles by its coefficient from the balanced equation.  The smaller of these two numbers is the limiting reactant.  Use the number and unit associated with the limiting reactant to solve the problem.

 

EX:  (A) How many grams of Fe2O3 can be produced from the reaction of 20.0 grams of Fe with 20.0 grams of O2?  (B) How many grams of excess reactant remains after reaction is complete?
                                    _4_ Fe  +  _3_ O2  à  _2_ Fe2O3


(A) STEP 1:  determine the limiting reactant

Fe: 20.0 g Fe | 1 mole Fe   = 0.358 moles Fe                   O2: 20.0 g O2 | 1 mole O2 = 0.625 moles O2         
                    | 55.85 g Fe                                                                     | 32 g O2

Fe: 0.358 = 0.0895                                                        O2: 0.625 = 0.208
         4                                                                                 3

So, Fe is limiting reactant.  Use the information given for Fe to solve the problem.

 

0.358 moles Fe = x moles Fe2O3               4x = 0.716

         4                          2                      x = 0.179 moles Fe2O3

 

0.179 moles Fe2O3 | 159.7 g Fe2O3 = 28.6 grams Fe2O3
                            | 1 mole Fe2O3


(B) STEP 1:  Using limiting reactant information, determine the amount of excess reactant used.
0.358 moles Fe  =   x moles O2                4 x = 1.074
          4                        3                         x = 0.2685 moles O2 used
0.2685 moles O2 | 32 g O2   =      8.59 g O2 used
                        | 1 mole
STEP 2:  Subtract number of grams of excess (O2) used from the original amount given in the problem.

20.0 g – 8.59 g = 11.41 g O2 remain

 

On your own…  if 25.00 grams of aluminum sulfate reacted with 25.00 grams of zinc chloride, how many grams of aluminum chloride could be produced?  How many grams of excess reactant remains after reaction is complete?

___ Al2(SO4)3  +  ___ ZnCl2  à  ___ AlCl3  +  ___ ZnSO4



 

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