# 4.2: Limiting & Excess Reagents

Learning Objectives

• Understanding Stoichiometric Proportions
• Understanding Limiting and Excess Reagents
• Predict quantities of products produced or reactants consumed based on complete consumption of limiting reagent (on both mole and mass basis)
• Predict quantities of excess reagents left over after complete consumption of limiting reagents.

## Stoichiometric Proportions and Theoretical Yield

A balanced chemical equation describe the ratios at which products and reactants are respectively produced and consumed. That said, the coefficients of the balanced equation have nothing to do with the actual quantity of reactants you start with, as you can mix any amount you choose, but clearly the maximum yield (theoretical yield) must be limited by the reactant that gets consumed up first, the limiting reagent. Any reagents remaining after the complete consumption of the limiting reagent are know as excess reagents.

Here is some common terminology used to describe reactions based on the concentrations of reactions.

• Stoichiometric Proportions: Reactants are mixed in the ratios defined by their stoichiometric coefficients. If a reaction proceeds to completion, everything is consumed.
• NonStoichiometric Proportions: Reactants are mixed in ratios that are different than the stoichiometric coefficients. One species runs out first (Limiting Reagent), while another is not completely consumed (Excess Reagent).
• Excess Reagent: The quantity (mole or mass) left over after the complete consumption of the limiting reagent

Quantity Excess = Initial Quantity - Consumed Quantity.

• Theoretical Yield: the maximum possible yield based on the complete consumption of the limiting reagent

Thought Question $$\PageIndex{1}$$

Are the limiting reagents always completely consumed?

No, only if the reaction goes to completion. There can be many different reasons why the limiting reagent is not completely consumed, these can include:

• Products also react to form reactants causing an equilibrium of reactants of products to coexist, this will be covered next semester (see Chapter 15)
• Reactions may not be over (some reactions occur very slowly).

## Limiting Reagent Problem Strategies:

1. Identify moles of all reactants present.
If given mass, divide by formula weight to convert to moles (this is the mass to mole step from the section 4.1,3.
2. Divide moles of each reactant by it's stoichiometric coefficient.
This is the denominator of the mole-to-mole step in section 4.1.3.
3. Smallest number indicates limiting reagent.
4. Multiply by stoichiometric coefficient of species you are solving for, and answer the question .
This is the numerator of the mole-to-mole step in section 4.1.3. If you are after moles, you are finished, if you are after mass, you need to use the molar mass of product to convert moles product to grams mass product, which is the mass-to-mole step in section 4.1.

### Limiting Reagent Problem

Silver tarnishes in the presence of hydrogen sulfide and oxygen due to the following reaction

4Ag + 2H2S + O2 ----> 2Ag2S + 2H2O

What is the Limiting Reagent and Theoretical Yield of Ag2S if 2.4 g Ag, 0.48 g H2S and 0.16g O2 react?

Note in the video how we first wrote the balanced equation, and then under each species wrote down what we were given. In this problem there are 3 reagents, and this technique allows us to quickly identify the

Now calculate the excess reagents for the reaction of 2.4 g Ag, 0.48 g H2S and 0.16g O2 to form Ag2S + H2O.

Exercise $$\PageIndex{1}$$

Consider the oxidation of glucose through respiration:

$C_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2+6H_2O + Energy$

What mass of carbon dioxide forms when 25.00 g of glucose reacts with 40.0 g of oxygen?

Step 1: Calculate moles of each reactant:

$$\mathrm{25.00\:g \times \dfrac{1\: mol}{180.06\:g} = 0.1388\: mol\: C_6H_{12}O_6}$$

$$\mathrm{40.0\:g \times \dfrac{1\: mol}{32\:g} = 1.25\: mol\: O_2}$$

Step 2: Divide each by its stoichiometric coefficient, smallest value is limiting reagent.

$1.25 mol O_2(\frac{1}{6mol})=0.208 \\ 0.1388 mol C_6H_{12}O_6(\frac{1}{1mol})=0.1388$

So glucose is the limiting reagent.

Step 3: calculate the mass carbon dioxide based on the complete consumption of the limiting reagent.

$0.1388 mol \; C_6H_{12}O_6(\frac{6mol \; CO_2}{1mol \; C_6H_{12}O_6})\left ( \frac{44.011g\; CO_2}{mol} \right )=36.66g \; CO_2$