1.    Write the balanced equation (reaction).

2.    Convert the known information (e.g. mass) of the reactant to moles.

        A.   If need help on determining the appropriate equation,
                look at Graphic Organizer.

        B.  Will modify this step when we get to Limiting Reactant problems

3.    Use the balance equation to set-up the right "Molar Ratio" to go from
        the known substance to the unknown substance.

4.    Convert from the moles of the unknown substance to the unknown information
        (e.g. are you looking for mass, volume, # of particles).

        A.   Again, if you need help on determining the appropriate equation,
               look at the Graphical Organizer.
 
 
 

Example:    Calculate the mass (in grams) of NH₃ (ammonia) produced by
                 the reaction of 5.40 grams of hydrogen with nitrogen.  The
                 balanced reaction equation is :
                                    1 N_2(g) +  3 H_2(g) =>  2 NH_3(g)

Step 1    Question (see balanced reaction above) gave is information

Step 2    Convert the mass of hydrogen (known substance) to moles

            #mole =    grams
                        molar mass                            Molar Mass of H₂
                                                                    (1g)(2) = 2g/mol

            #mole = 5.40g H₂   =   2.70 mole H₂
                        2g/mol H₂

Step 3    Figure out from the balanced reaction what the Molar Ratio is
              between hydrogen and ammonia.

               2 mole NH₃
                3 mole H₂

Now use this ratio to convert from known (hydrogen) to unknown (ammonia)

        ( 2.70 mole H₂ )(    2 mole NH₃ )   =  1.80 mole NH₃
                                    3 mole H₂

Step 4    Convert from unknown (ammonia) moles to unknown information (mass).

            #mole =    grams
                        molar mass                            Molar Mass of NH₃
                                                                    (14g)(1) + (1g)(3) = 17g/mol

         1.80 mole NH₃ =       X
                                   17g/mol NH₃

        X = ( 1.80 mole NH₃ )(17g/mol NH₃) = 30.60 grams NH₃

        X =  30.6 grams NH₃