Now that we know how to relate quantities of atoms, elements or molecules to real-world measurable quantities such as grams, it is time to combine our knowledge of quantities and chemical equations to plan and predict chemical reactions.



Given the following balanced equation:

Mg(OH)2 + 2HCl ® MgCl2 + 2H2O We know that exactly 1 mole of Mg(OH)2 will react exactly with just 2 moles of HCl to produce exactly 1 mole of MgCl2 and exactly 2 moles of H2O. Consequently, if we know exactly how many moles of Mg(OH)2 we have available, we can calculate exactly how many moles of HCl are required to react with it. Calculate the number of grams necessary to react completely with 5.02 g of Mg(OH)2

initial mass of A                         MM of A              mole ratio B:A             MM of B              calc. mass of B
5.02 g Mg(OH)2      x      1 mol    x     2 mol HCl    x     36.5 g    =     6.29 g HCl
                                       58.3g         1 mol Mg(OH)2     1 mol

The mole ratio of species comes from the ratio of the coefficients of the species in the balanced chemical equation and is used to convert between moles of different species that occur in a chemical equation (note: the mole ratio can only be used with moles, never, never with grams of a substance).



Try this problem:

You have a 9.00 g mixture of KClO3 and rust. Determine the %KClO3 in the mixture if the mixture loses 2.00 g upon heating.

                                                    D                                                (remember D means heat)
                        2KClO3(s) ® 2KCl(s) + 3O2(g)



Note that in the aluminum problem above, we specified a given amount of one reactant, reactant A, and indicated an unspecified amount of the other reactant. The "required amount", or the stoichiometric amount, is that amount which would react exactly with reactant A as specified by the mole ratio in the balanced chemical equation. An "excess amount" is just that, more than what is exactly needed for the reaction.

In some problems you will be given non-stoichiometric amounts of both reactants and you will have to determine which is the limiting reactant or the reactant which is present in the smaller stoichiometric amount.

for example:

You have 10 bicycle frames and 16 bicycle wheels and you need to put them together to produce as many bicycles as possible, how many bicycles can be produced, what is your limiting "reactant", and how much excess "reactant" do you have left over?

Balanced equation:

                       one frame + two wheels ® one bicycles
                              10 frames           16 wheels                8 bicycles

the only way to produce one bicycle is with a ratio of:      two wheels
                                                                                            one frame

Is:         16 wheels the same ratio, larger ratio, or smaller ratio than 2 wheels?
             10 frames                                                                               1 frame

The ratio is smaller so the wheels are the limiting "reactant" (the wheels are stoichiometrically fewer in number than the frames) and the frames are the excess reagent. So how many bicycles can be produced? We have enough wheels for 8 bicycles but enough frames for 10 bicycles. It is obvious that we can only make as many bicycles as we have wheels for, or 8 bicycles.

Always use the limiting reactant for stoichiometric calculations:

16 wheels     x     1 bicycle     =       8 bicycles
                            2 wheels

8 bicycles    x       1 frame      =       8 frames
                            1 bicycle

10 frames - 8 frames    =    2 frames are left over

The stoichiometric quantities for this problem are 8 frames and all 16 wheels produce 8 bicycles – the other two frames are useless because there are no wheels available to produce a bicycle.




The Theoretical Yield is the maximum amount of product that can be obtained from a reaction given the specified amounts of the reactants. In other words, the Theoretical Yield is just the calculated yield of product based upon the limiting reactant.

Few reactions will actually produce the maximum amount of product that can be made due to experimental error, incomplete reaction, poor technique in isolating the product, etc. Therefore we calculate the %Yield to give us some indication as to how complete the reaction is and how good our experimental technique is.

The Percent Yield       =         actual yield      x       100
                                           theoretical yield


Tips for Stoichiometry Problems:

You must start with a balanced equation in order to recognize the correct "mole ratio" between species--since these are your conversion factors:

                          2H2 + O2 ® 2H2O

The mole ratio is:         2 mole H2          in conversion factor form
                                     1 mole O2

                                     2 mole H2O
                                     1 mole O2   

                                     2 mole H2O
                                     2 mole H2

The mole ratio converts between moles of "different substances" in the balanced chemical equation

The pattern for solving stoichiometry problems is:

    start           convert using                           convert using                        convert using            end
    with           molar mass A                          mole ratio B:A                         molar mass            with
grams A             ®            moles A          ®            moles B         ®           grams B

If the problem is a limiting reactant problem, you must first determine which reactant is the limiting reactant, in other words, which reactant is present in the smallest stoichiometric amount or will be used up completely in the reaction.

All stoichiometric calculations will be based only on the initial limiting reactant amount, in other words, start all calculation procedures with the limiting reactant amount (whether it be in grams or moles). In other words, A (in the above scheme) is always the limiting reactant.

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