Chapter 12 Notes

Derived from the Greek words:

    stoicheion meaning “element”

    metron meaning “measure”

The relationship between quantities (mass of substance or volume of gas)in a chemical reaction according to the chemical equation.

12.1 The Arithmetic of Equations

Key Concepts:1. How is a balanced equation like a recipe?2. How do chemists use balanced chemical equations?3. In terms of what quantities can you interpret a balanced chemical equation?4. What quantities are conserved in every chemical reaction?

Using everyday equations:- A balanced equation provides the same kind of quantitative information that a recipe does.

Using Balanced Chemical Equations:- Chemists use balanced chemical equations as a basis to calculate how much reactant is needed or product is formed in a reaction.

The equation must be balanced properly to produce the right product, but what it is really showing is the exact ingredients necessary to make the product properly.

Interpreting Chemical Equations

-Number of atoms-Number of molecules-Moles-Mass-Volume-Mass conservation-Mole Ratio!!!!!!!!!

12.2 Chemical Calculations:

1. How are mole ratios used in chemical calculations?2. What are the general types of stochiometric equations?3. What are the basic steps for solving stochiometric equations?

Types of Stoichiometric Problems

List the possibilities!

Steps to solve problems:

12.3 Limiting Reagent and Percent Yield:

1. How is the amount of product in a reaction affected by an insufficient quantity of any of the reactant?

2. What does the percent yield of a reaction measure?

Limiting Reagent – In a chemical reaction, an insufficient quantity of any of the reactants will limit the amount of product formed. This reactant is know as the limiting reagent and the other reactant(s) are excess reagents.

Steps:

1. Change the given amount of each reactant to moles.2. Multiply that value by the mole-mole ratio. (cancel like values)3. Compare calculated values to given values.4. The limiting reagent is the calculated value that is less than the

given value.

A 50.6 g sample of Mg(OH)2 is reacted with 45.0 g of HCl according to the reaction:

Notice how quantities of both reactants are known. Which one will be used up first? You can't tell, nor should you jump to any conclusions. Just because it looks like there is less Mg(OH)2 present does not automatically mean it will be used up before all of the HCl is consumed. This is a limiting reagent problem.

In order to find out which reactant is the limiting reagent, you have to compare them to each other. This comparison must be done in moles, therefore, the next step will be to convert each of the grams of reactants to moles:

Again, you should not jump any conclusions about which reactant is the limiting reagent. Just because there are fewer moels of magnesium hydroxide does not mean it is the limiting reagent. Arbitrarily pick one of these reactants and calculate how many moles of the other reactant is needed to completely use up the reactant picked. In this case, magnesium hydroxide is arbitrarily chosen:

Compare the mols HCl needed to the actual mols HCl available. In this case, 1.74 mol of HCl is needed and 1.23 mol HCl is available--that's not enough. So, even though it appears that there are more mols of HCl than Mg(OH)2, the HCl is the limiting reagent. The HCl will be run out before the magnesium hydroxide and thereby limit the amount of product formed.

For this reason, use the mols of HCl to calculate the theoretical yield of magnesium chloride:

The theoretical yield is the maximum amount of product which can be produced (in an ideal world). In the "real" world it is difficult to produce the amount obtained for the theoretical yield.

A percent yield is often used to show how close to ideality one has obtained in a chemical synthesis. Suppose in the reaction discussed a chemist actually obtained 55.4 g of MgCl2. This is called the actual yield and would be given to you in the problem. To calculate the percent yield:

Slides 11-17 http://pages.towson.edu/ladon/limreas.html