The Limiting Reagent

The chemical calculations we've worked out thus far have assumed that there was an excess of reactants. In other words we did not have to worry about one or more reactants running out before the reaction was complete. For example, consider the generaic reaction:

A + 2B --- > AB2

If we begin with 1 mole of A and 1 mole of B how many moles of AB2 will we get? The reaction stoichiometry states that 1 mole of A will react with 2 moles of B to produce 1 mole of AB2. Since we need 2 moles of B to completely react with 1 mole of A, we will run out of B first (there is only one mole of B available). Whatever reactant runs out first is called the limiting reactant or limiting reagent. Thus B is the limiting reagent in the above scenerio. The reaction stops after the limiting reagent runs out. In the above case 1 mole of B reacts with 0.5 mol of A to produce 0.5 mole of AB2 with 0.5 mol of A unreacted.

Let's consider a specific reaction. Consider the reaction of 80 g of copper with 50 g of molecular oxygen. Let's answer the following questions: (1) What is the limiting reagent? (2) What mass of product is formed? (3) What is the mass of excess reactant? To answere these questions we need the balanced chemical equation:

1) Determine moles of each reactant:

2) Choose one reactant (it does not matter which) and determine how many moles of the other reactant are necessary to completely react with it. Let's choose Cu: How many moles of O2 are necessary to completely react with 1.26 mol of Cu? This part of the calculation is done with the mol-mol conversion factors:

3) Use this information to determine the limiting reagent:

The calculation above means that we need 0.63 mol of O2 to completely react with the copper. We have 1.56 mol of O2 and therefore more than enough oxygen. Thus oxygen is in excess and copper must be the limiting reagent.

4) Mass of product: Since the limiting reagent determins how much product is formed we use the moles of limiting reagent to determine the moles and mass of product:

5) Excess mass of reactant:

!.56 moles of O2 was initially present and 0.63 moles were consumed in the reaction. Thus 1.56 - 0.63 = 0.93 mol of O2 was in excess. 0.93 mole of O2 is equal to 29.8 g of O2. In other words 29.8 g of O2 was in excess.

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C101 Class Notes
Prof. N. De Leon