1 / 12

Stoichiometry

Learn about the concepts of stoichiometry in chemistry, including the accounting of chemistry through moles, the use of molar mass in understanding quantities in compounds and reactions, and how to use moles and molar mass in a chemical reaction. Explore the determination of limiting reactants and the production of products, as well as the application of stoichiometry principles to quantities of gases and solutions.

winonaw
Download Presentation

Stoichiometry

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Stoichiometry The accounting of chemistry

  2. Moles • What are moles? • Moles are a measure of matter in chemistry. • Moles help us understand what happens in a chemical reaction.

  3. Molar mass • What do we use molar mass for? • Molar mass, along with moles, helps us understand quantities in compounds and reactions. • Ex. How many grams are found in 1/3 mole of KClO3 ?

  4. Stoichiometry • How do we use moles and molar mass in a chemical reaction? • Moles and molar mass are used in ratios to show how much of each compound is needed, and how much will be produced.

  5. Stoichiometry • How do we use this? • First we need a correctly written and balanced equation. • Ex. 23.8 grams of potassium iodide crystals reacts with 34.7 grams of chlorine gas (diatomic) to make potassium chloride crystals and iodine gas (diatomic).

  6. Stoichiometry • 2KI + Cl2 2KCl + I2 • What are the molar masses? • KI = 166.00 g/mol • Cl2= 70.906 g/mol • KCl = 74.551 g/mol • I2 = 253.80 g/mol • KI = 23.8g/166.00g/mol = .143 mol • Cl2 = 34.7g/70.906 g/mol = .489 mol

  7. Stoichiometry • 2KI + Cl2 2KCl + I2 • Using the ratios from the balanced equation, we can use the molar mass to determine the limiting reactant. • 2 mol KI for every 1 mol of Cl2 • .143 mol KI x 1 Cl2/2 KI = .0715 mol Cl2 • .489 mol Cl2 x 2 KI/1 Cl2 = .979 mol KI

  8. Stoichiometry • What does this mean? • You need .0715 mol of Chlorine gas but you have .489 mol. You have way too much (excess reactant) chlorine. • You need .979 mol of KI but only have .143, that is not enough (limiting reactant) KI.

  9. Stoichiometry • How much Iodine is made? • 2KI + Cl2 2KCl + I2 • Again use the mole ratio and the limiting reactant. • .143 mol KI x 1 I2/2 KI = .0715 mol I2 • .0715 mol I2 x 253.80 g/mol = 18.1 g • 10.7 g of KCl

  10. Stoichiometry • How can we do this same process with quantities of gases? • Avogadro’s principle states that equal volumes of gases and the same pressure and temperature will have equal numbers of particles. • 22.4 Liters = 1 mole (at STP)

  11. Stoichiometry • If 90 L of oxygen and 60 L of methane (CH4) react at a constant temperature and pressure (STP), which is the limiting reactant, and how much carbon dioxide could be made?

  12. Stoichiometry • What about with solutions? • Solutions have a given concentration. • The concentration “molar” is moles per liter. • How many moles of HCl are in 25 mL of .10 molar solution?

More Related