Limiting Reagent and Theoretical Yield

Section 3

Limiting and Excess Reagents

Section 3 – Part 1

Determine the limiting reagent in a reaction Calculate the amount of product formed in a

reaction, based on the limiting reagent. Calculate the amount of excess reagent left

over in a reaction.

Objectives

If you are given one dozen loaves of bread, a gallon of mustard and three pieces of salami, how many salami sandwiches can you make?◦ The number of sandwiches you can make will be

limited by the amount of salami you have, so the salami is the limiting ingredient.

The limiting reagent is the reactant you run out of first.

The limiting reagent determines the amount of product you can make

When all of this reactant is used up, the reaction will stop, and no more product will form.

Limiting Reagent

The excess reagent is the reactant that is not completely used up in the reaction.

Some of this reactant will remain unreacted when the reaction stops.

Excess Reagent

You must do two stoichiometry problems. To determine the limiting reagent:

1. Convert the amount of each reactant to the amount of product. This will be a 1, 2, or 3 step stoichiometry problem for each reactant.

2. The reactant that produces the LEAST amount of product is the limiting reagent.

3. The other reactant is the excess reagent.

** Note: You can convert the amount of reactants to either moles or grams of product. Moles is sufficient, but if you are asked to determine grams of product, converting to grams in one conversion problem is easiest.**

Determining the Limiting Reagent

Copper reacts with sulfur to form copper(I) sulfide. If 10.6 g of copper reacts with 3.83 g S, what is the limiting reagent?

First: Write the balanced equation for the reaction.

Example

Then: Convert grams of copper to moles of product and grams of sulfur to moles of product (2 stoichiometry problems).

Example (continued)

Determine which reactant produces the LEAST amount of product. This is the limiting reagent.

The other reactant is the excess reagent.

Example (continued)

Example:If 10.6 g of copper reacts with 3.83 g S, how many grams of product will be formed?

This time, we will convert to grams of product since we are asked for that.

Using Limiting Reagent to Find Quantity of Product

If 10.6 g of copper reacts with 3.83 g S, how many grams of product will be formed?

2Cu + S ® Cu2S

10.6 g Cu 63.55g Cu 1 mol Cu

2 mol Cu 1 mol Cu2S

1 mol Cu2S

159.16 g Cu2S

= 13.3 g Cu2S

3.83 g S 32.06g S 1 mol S

1 mol S 1 mol Cu2S

1 mol Cu2S

159.16 g Cu2S

= 19.0 g Cu2S

= 13.3 g Cu2S

Cu is Limiting Reagent

We can determine the amount of excess reagent that is left over.

Use the limiting reagent to find out how much excess reagent you used in the reaction.

Subtract the amount used from the amount of excess reagent you started with to get the amount left over.

Determining the Amount of Excess Reagent

If 4.87 mol of magnesium and 9.84 mol of HCl are reacted, what is the limiting reagent?

Mg(s) +2 HCl(aq) ® MgCl2(s) +H2(g) How many moles of gas will be produced? How much excess reagent remains?

Example:

If 10.3 g of aluminum are reacted with 51.7 g of CuSO4 how much copper will be produced?

How much excess reagent will remain?

Example