Stoichiometry is where we make the connection between balanced equations and moles. In the laboratory, we can only measure masses of various substances. Equations specify how many atoms or molecules are involved in a reaction.

It is valid to interpret coefficients in terms of moles. For example, if an equation specifies one Ca combines with two HCl's, we can assume a 1:2 ratio of moles. Since we know how mass and moles are related, we can now make the connection between massing in the lab and calculations using equations.

Making Predictions in Chemical Reactions

If we know how much of a reactant we have, we can predict what amount of product will be formed. If we start with 20.0 g of Ca and excess HCl, how much CaCl2 is formed?

• Step 1: Write the balanced equation

Ca + 2 HCl --> CaCl2 + H2

• Step 2: Convert mass of starting material to moles

                                              20.0 g Ca | 1 mole Ca = 0.50 mole Ca
                                                                | 40 g Ca

• Step 3 Set up a Before: Change: After (BCA) table

Ca       +     2 HCl   -->    CaCl2       +           H2

Before                           0.5 moles       XS                 0                          0

Change                        - 0.5 moles       ?                    ?                          ?

After

• Step 4: Determine how much of the other reactant is consumed and how much of the products are formed by using coefficients as guides. Then fill in BCA table.

moles HCl = 2x moles Ca

moles CaCl2 = moles Ca

moles H2 = moles Ca

Ca       +     2 HCl   -->    CaCl2       +           H2

Before                           0.5 moles       XS                 0                          0

Change                        - 0.5 moles     -1mole      + 0.5 moles         + 0.5 moles      

After                                  0 moles        1 mole      0.5 moles            0.5 moles

• Step 5: Convert moles of required to mass of required

                              0.5 moles CaCl2 | 111.0 g CaCl2 = 55.5 g CaCl2
                                                            | 1 mole CaCl2

Some Flies In the Ointment: Limiting Reactants

Whenever you cook, you follow a recipe. For example with pizza, you need pizza dough, sauce and cheese. If you run out of pizza dough, you cannot make any more pizza, no matter how much sauce and cheese you have left. The pizza dough is the limiting factor in how many pizzas you can make.

The same idea holds for reactions.Sometimes you are given quantities of both reactants. In this case you have to determine which one runs out first. This is the limiting reactant and determines how much product is formed. Once the limiting reactant is consumed, the reaction stops.

Let's use the above reaction as an example. If we start with 20.0 g of Ca and 20.0 g of HCl, which reactant will be limiting and how much CaCl₂ will be formed? We start the same way.

• Step 1: Write the balanced equation

Ca + 2 HCl -->  CaCl₂ + H₂

• Step 2: Convert mass of starting material to moles

                                           20.0 g Ca | 1 mole Ca = 0.50 mole Ca
                                                             | 40 g Ca

                                        20.0 g HCl | 1 mole HCl = 0.548 mole HCl
                                                            | 36.5 g HCl

• Step 3 Set up a Before: Change: After (BCA) table

Ca      +        2 HCl   -->          CaCl₂          +             H₂

Before                  0.5 moles      0.548 moles           0                                0    

Change

After

• Step 4: Determine which reactant is in excess and which is limiting by comparing what you have with what you need.

moles HCl = 2x moles Ca

We have 0.5 moles Ca and need 1.0 moles of HCl. Since we only have 0.548 moles of HCl, it will be the limiting reactant and determine how much Ca is consumed and how much CaCl2 is formed.

• Step 5: Set up BCA table

Ca        +        2 HCl     -->          CaCl₂     +             H₂

Before                       0.5 moles         0.548                       0                       0

Change                - 0.274 moles    -0.548 moles         +0.274 moles   + 0.274 moles

After                        0.225 moles    0 moles                   0.274 moles      0.274 moles

• Step 6: Convert moles of required to mass of required

                           0.274 moles CaCl2 | 111.0 g CaCl₂ = 30.4 g CaCl₂