STOICHIOMETRY PROBLEMS

 

1.         STOICHIOMETRY IS THE STUDY OF THE RELATIONSHIP OF QUANTITIES OF REACTANTS USED TO PRODUCTS PRODUCED.  IN OTHER WORDS, IF I NEED 20 GRAMS OF HYDROGEN, HOW MUCH MAGNESIUM DO I NEED TO MAKE IT?

 

2.         WHEN YOU LOOK AT A BALANCED EQUATION, YOU HAVE BEEN TAUGHT TO THINK IN NUMBERS OF ATOMS OR MOLECULES BUT THE SAME RELATIONSHIP EXISTS FOR MOLES.

 

            EXAMPLE:     CaCO3     =    CaO     ÷    CO2

 

THIS COULD BE READ AS 1 MOLECULE OF CaCO3 FORMS 1 MOLECULE OF CaO PLUS 1 MOLECULE OF CO2.  HOWEVER, JUST AS EASILY IT COULD BE READ 1 MOLE OF CaCO3 MOLECULES FORMS 1 MOLE OF CaO MOLECULES PLUS 1 MOLE OF CO2.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

MOLE MOLE PROBLEMS

 

IF YOU HAD 6 MOLES OF WATER HOW MANY MOLE OF OXYGEN WOULD BE FORMED WHEN ELECTRICITY WAS RUN THROUGH THE SOLUTION.

 

                        STEP 1 -   WRITE THE BALANCED EQUATION

                                    2H2O     =     2H2    +    O2

 

STEP 2 – OUR BALANCED EQUATION SHOWS THAT 2 MOLES OF H2O FORM 1 MOLE OF OXYGEN.  THIS SAME RELATIONSHIP EXISTS FOR ANY NUMBER OF MOLES.  THIS BECOMES A SIMPLE CONVERSION.

           

            6 MOLES OF H2O      x     1MOLE OF O2    =   3 MOLES OF

                                                         2MOLES OF H2O       OXYGEN

 

THE LAW OF CONSERVATION OF MATTER CAN BE PROVEN BY THE USE OF MOLES.  FOR EXAMPLE IN THE EQUATION OF FOMATION OF WATER THE MASS OF THE REACTANTS AND PRODUCTS CAN BE FOUND BY USING MOLES

 

                        2 H20        =     2H2 + O2

 

JUST CONVERT THE MOLES OF EACH TO GRAMS AND ADD THE AMOUNTS TOGETHER.  THE GRAMS OF REACTANTS SHOULD EQUAL THE GRAMS OF PRODUCTS.  TRY IT.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

SOLVING STOICHIOMETRY PROBLEMS

 

1.         THERE ARE THREE MAJOR CATOGORIZE OF THESE PROBLEMS, TO DO THEM YOU WILL NEED TO BE ABLE TO RECOGNIZE EACH TYPE.

 

                        A.        MASS TO MASS

                        B.         MASS TO VOLUME

                        C.        VOLUME TO VOLUME

 

2.         THE MASS TO MASS PROBLEM IS JUST ONE STEP BEYOND THE MOLE TO MOLE PROBLEMS YOU HAVE ALREADY DONE.

 

EXAMPLE:     IF YOU HAVE 54 GRAMS OF H2O, HOW MANY GRAMS OF OXYGEN WILL BE FORMED WHEN ELECTRICY IS RUN THROUGH THE WATER?

 

STEP 1 – WRITE THE BALANCED EQUATION

 

            2H2O    =    2H2   + O2

 

THE STEPS OF STOICHIOMETRY PROBLEMS FOLLOWS THE SAME STEPS OF PORTIONS OF THE STEPS AS SHOWN BELOW.  EACH STEP IS MOVED THROUGH BY USING CONVERSION FACTORS.

 

                        54GMS                                                MASS 48GMS

           

     1 MOLE = 18GMS  2H2O           =       2H2   +   O2   1.5MOLES   x   32 GMS/MOLE

 

            54GMS/18GMS = 3 MOLE                            MOLE 1.5

 

                        3 MOLES OF H2O    x   1MOLE O2  = 1.5 MOLES

                                                                 2 MOLE H2O

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2.         THIS SAME PATTERN IS USED FOR MASS VOLUME PROBLEMS BUT USING THE APPROPRIATE CONVERSION FACTOR.  MEMORIZE THIS PATTERN SO WHEN YOU READ A PROBLEM YOU WILL KNOW EXACTLY WHERE YOU ARE AT AND IT WILL FIT A LOGICAL PATTERN THAT CAN BE LEARNED AND NOT MEMORIZED.  IF YOU MEMORIZE WITHOUT UNDERSTANDING YOU WILL BE LOST THE REST OF THE YEAR.

 

3.         VOLUME-VOLUME PROBLEMS ARE SIMPLER BECAUSE THE VOLUME IS ALWAYS 22.4L FOR A MOLE.  THEREFORE, YOU DON’T HAVE TO GO FROM VOLUME TO MOLES THEN BACK TO VOLUME.  (THIS WOULD JUST DIVIDE IN 22.4 AND DIVIDE IT BACK OUT.)  YOU CAN GO DIRECTLY FROM VOLUME TO VOLUME USING THE MOLE RATIO OF THE BALANCED FORMULA.

 

            EXAMPLE:    WHEN WATER WAS BROKEN INTO HYDROGEN AND OXYGEN.  44.8 LITERS OF OXYGEN WAS PRODUCED.  HOW MANY LITERS OF HYDROGEN WAS PRODUCED?

 


                        44.8 L  H2VOL.   x   1.02/2H2 = 22.4L VOL

 


            2 H2O           =       2H2          ÷         O2

 

4.         NUMBER OF PARTICLES COULD BE USED IN PLACE OF MASS OR VOLUME BUT THE MECHANICS ARE ALL THE SAME AS BEFORE.  YOU JUST USE THE CONVERSION FACTOR OF 1 MOLE = 6.023   x   10 INSTEAD.

 

5.         ANOTHER POSSIBLE VERSION COULD BE STARTING AT MOLES AND GOING TO SOME OTHER MEASUREMENT.  AGAIN THE MECHANICS ARE THE SAME YOU JUST HAVE ONE LESS STEP IS ALL.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

LIMITING REACTANTS AND PERCENT YIELD

 

 

1.         WHEN TWO REACTANT ARE INVOLVED THE ONE IS LEAST AMOUNT WILL STOP THE REACTION.  YOU CAN’T TELL THIS BY WEIGHT SO YOU HAVE TO CONVERT TO MOLES AND SEE WHICH IS IN LEAST AMOUNT BY SEEING HOW MUCH OF THE OTHER REACTANT IS NEEDED.

 

            EXAMPLE:   IF 16 GRAMS OF OXYGEN REACTED WITH 14 GRAMS OF HYDROGEN HOW MUCH WATER WOULD BE FORMED?

 

            TO FIND THE REACTANT THAT RUNS OUT FIRST FIND THE NUMBER OF MOLES OF EACH AND CONVERT TO THE AMOUNT OF THE OTHER NEEDED.

 

                                                2H2O        =      2H2    +     O2

               H2   =   14/2GRMS   =   7 MOLES   x   2/2x32GRM/1   =  112 GMS OR O2

           O2   =   16 GMS/32GMS   =   1/2 MOLE x 2/1   x 2 GRAMS = 2 GRAMS OF H2

 

SINCE 15 GRAMS OF O2 ONLY NEED 4 GRAMS OF H2 AND 12 GRAMS OF H2 WOULD NEED 118 GRAMS OF O2 TO COMPLETE THE OXYGEN WOULD RUN OUT FIRST.  THEREFORE, TO CALCULATE THE AMOUNT OF WATER FORMED THE OXYGEN IS USED IN THE MASS-MASS PROBLEM INSTEAD OF HYDROGEN.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2.         IN A PERCENT YIELD PROBLEM, THE PERCENT IS FOUND AS IN ANY PERCENT PROBLEM.  WHEN A CHEMICAL REACTION OCCURS OFTEN THINGS DON’T HAPPEN PERFECTLY AND THERE IS NOT AS MUCH PRODUCTS AS THE CACULATIONS SAY IT WOULD BEE, IF CONDITIONS WERE PERFECT.  THEREFORE, THE ACTUAL YIELD IS JUST DIVIDED BY THE EXPECTED YIELD AND MULTIPLIED BY 100.

 

                        PERCENT YIELD   =   ACTUAL YIELD      x      100

                                                                EXPECTED YIELD