Continuing Stoichiometry…

The idea.  In every chemical reaction, there is one reactant that will be run out (called the limiting reactant).  This will limit the amount of product that can form.  The reaction will stop at that point.  There are then also excess reactants.

Sample Problem  S Cl 2  4 S 2 Cl 2 If you have g S 8 and g Cl 2, what is the limiting reactant?

Step 1  First, find the moles of each reactant.  Amount of each reactant will be given  Usually it will be in grams. This is a gram to mole conversion.

The Answers g S 8 1 mole S 8 x =.78 mol S g S g Cl 2 1 mole Cl 2 x = 1.41 mol Cl g Cl 2

If moles are Given…  If moles are given, problem starts here…

Step 2: Compare Mole Ratios  A) What is the mole to mole ratio from the balanced chemical equation?  S Cl 2  4 S 2 Cl 2 Moles S 8 = 1 Moles Cl 2 = 4 Need 4 Cl 2 for each S 8 !

Step 2 cont’d  B) What is the mole to mole ratio calculated from the starting conditions? Moles S 8 =.78 mol S 8 Moles Cl 2 = 1.41 mol Cl mol Cl 2 = 1.81 mol Cl 2 for 1 mol S 8.78 mol S 8

Is this enough?  NO!  Therefore, Cl 2 is the limiting reactant Now practice some LR calculations!

Calculating the Amount of Product Formed  Complete a mole to gram calculation using the Limiting Reactant  In our example:  LR: Cl 2 = 1.41 moles Try it!

The Answer  S Cl 2  4 S 2 Cl mol Cl 2 4 mol S 2 Cl g S 2 Cl 2 x x 1 4 mol Cl 2 1 mol S 2 Cl 2 = g S 2 Cl 2

Analyzing the Excess Reactant  You care about two things:  How much of the excess reactant reacted  How much of the excess reactant remains

Excess that Reacted  Complete a mole to gram calculation using the limited and excess reactants  In our example:  Limiting Reactant: Cl2 = 1.41 mol  Excess Reactant: S 8

The Calculation  S Cl 2  4 S 2 Cl mol Cl 2 1 mol S g S 8 x x 1 4 mol Cl 2 1 mol S 8 = g S 8

Excess Remaining  Subtract amount that reacted from the amount that you started with  In example:  Amount Reacted = g S 8  Amount Starting With = g S 8

The Answer g g = g S 8 in excess