Stoichiometry Chapter 9 Limiting Reagents Stoich ppt _5 Limiting Reagents Practice.

Slides:

Advertisements

Similar presentations

Stoichiometry Ratios The stoichiometric coefficients in a balanced chemical reaction can be used to determine the mole relationships between any combination.

Advertisements

Reaction Stoichiometry Chapter 9. Reaction Stoichiometry Reaction stoichiometry – calculations of amounts of reactants and products of a chemical reaction.

Limiting Reagent What happens in a chemical reaction, if there is an insufficient amount of one reactant?

Section 3: Limiting Reactants

 CHEM.B Apply the mole concept to representative particles (e.g., counting, determining mass of atoms, ions, molecules, and/or formula units). 

Limiting reagent, Excess reactant, Theoretical or Percent yield

Starter S moles NaC 2 H 3 O 2 are used in a reaction. How many grams is that?

Stoichiometry The Study of Quantitative Relationships.

In any chemical reaction, it is easy to run out of one or another reactant – which has an impact on the amount of products that can result from a reaction.

Information given by chemical equations

and cooking with chemicals

Stoichiometry Chapter 8. Stoichiometry Chemical equations Limiting reagent Problem types Percent yield Mass-mass Mole - mole other.

Stoichiometry Calculating Masses of Reactants and Products.

STOICHIOMETRY Chapter 9: Pages

Stoichiometric Calculations Stoichiometry – Ch. 9.

Stoichiometry. Information Given by the Chemical Equation  The coefficients in the balanced chemical equation show the molecules and mole ratio of the.

Proportional Relationships b Stoichiometry mass relationships between substances in a chemical reaction based on the mole ratio b Mole Ratio indicated.

Stoichiometry. The study of chemical changes is at the heart of chemistry. Stoichiometry is the area of study that examines the quantities of substances.

I. I.Stoichiometric Calculations Topic 9 Stoichiometry Topic 9 Stoichiometry.

I. I.Stoichiometric Calculations Stoichiometry. History of Stoichiometry b Comes from the Greek: Stoicheion - to measure the elements.

Chapter 12 Stoichiometr y. STOY-KEE-AHM-EH-TREE Founded by Jeremias Richter, a German chemist Greek orgin: stoikheion: element & metron: measure Stoichiometry—the.

Chemical Calculations Stoichiometry u OBJECTIVES: Calculate stoichiometric quantities from balanced chemical equations using units of moles, mass, representative.

Chapter 12 Notes. Information given by chemical equations 2 C 6 H 6 (l) + 15 O 2 (g)  12 CO 2 (g) + 6 H 2 O (g)  In this equation there are 2 molecules.

The Mole & Stoichiometry!

The Mole, part II Glencoe, Chapter 11 Sections 11.3 & 11.4.

Stoichiometry Chemical Quantities Chapter 9. What is stoichiometry? stoichiometry- method of determining the amounts of reactants needed to create a certain.

Continuing Stoichiometry…. The idea.  In every chemical reaction, there is one reactant that will be run out (called the limiting reactant).  This will.

1 Chapter 9 Mole Factors Calculations with Equations Limiting Reactions Percent Yield.

Calculate the mass of Cu produced? Mass of beaker and Cu – mass of beaker.

Define mole ratio (What is it? How is it determined?)

Chemistry 20 Stoichiometry. This unit involves very little that is new. You will merely be applying your knowledge of previous units to a new situation.

Stoichiometry is… Greek for “measuring elements” Defined as: calculations of the quantities in chemical reactions, based on a balanced equation. There.

Stoichiometry and cooking with chemicals.  Interpret a balanced equation in terms of moles, mass, and volume of gases.  Solve mole-mole problems given.

Can’t directly measure moles Measure units related to moles: –Mass (molar mass) –Number of particles (6.02 x ) –Liters of gas (22.4 Liters at STP)

Mass to Mass Conversions. Mole to Mole Conversions are the CRUCIAL LINK Follow the same steps: Step 1: Balance the equation Step 2: Write down what you.

Stoichiometry Chapter 9 Mole-to-mole ratios Stoich ppt _1 mole-mole.

Limiting Reagent If you have 20 wheels and 15 seats, how many bikes could you make? What will be in excess? How many pieces will be left over?

Stoichiometry GPS 13. Stoichiometry Example: 2H 2 + O 2 → 2H 2 O Equivalencies: 2 mol H 2 for every 1 mol O 2 2 mol H 2 for every 2 mol H 2 O 1 mol O.

Students type their answers here

Stoichiometry Chapter 9 Percent Yield Stoich ppt _5 Percent Yield.

Raider Rev Up Calculate moles in g of HgS (mercury II sulfide) MAR

Stoichiometry: Chapter 9 Lesson 1. Law of Conservation of Mass “We may lay it down as an incontestable axiom that, in all the operations of art and nature,

Stoichiometry Chapter 9 Limiting Reagents Stoich ppt _4 Limiting Reagents Intro.

Stoichiometry – Ch What would be produced if two pieces of bread and a slice of salami reacted together? + ?

1 1)Write correct chemical formulas, know types of reactions and write a balanced equation!!! 2)a) Convert g, L or # or particles to mole and b) use mole.

Stoichiometry Chapter 9 Mass calculations Stoich ppt _2 mass-mass.

 I have 5 eggs. How many cookies can I make? 3/4 c. brown sugar 1 tsp vanilla extract 2 eggs 2 c. chocolate chips Makes 5 dozen cookies. 2 1/4 c. flour.

DO NOW!!! Back of Worksheet!

Limiting Reagent What happens in a chemical reaction, if there is an insufficient amount of one reactant?

Stoichiometry The calculation of quantities in chemical equations.

Ch. 9 Notes -- Stoichiometry

7.4 Calculations Involving Limiting Reagents

Unit 8: Stoichiometry: Part 1

Chemical Reactions Unit

Unit 8 Stoichiometry Notes

Limiting Reactant/Reagent Problems

Reaction Stoichiometry

Unit 4 Stoichiometry Stoichiometry, mol-mol Stoich. , volume-volume

Chemical Reactions Unit

Information given by chemical equations

Stoichiometry How does stoichiometry relate to a correctly balanced chemical equation?

Stoichiometry Stoichiometry is a fancy chemistry word meaning “to calculate quantities in a chemical reaction”. Or in other words, if I mix this much of.

Chapter 12 Stoichiometry

Mole ratios Mole to mole conversions

Stoichiometry Follow along in your text Chapter 9 Section 1 Pages 302 – 311 Put on Pg.33 of your notebook!

STOICHIOMETRY!!!!.

Presentation transcript:

Stoichiometry Chapter 9 Limiting Reagents Stoich ppt _5 Limiting Reagents Practice

Stoichiometry After this presentation, you should understand:  Moles, mass, representative particles (atoms, molecules, formula units), molar mass, and Avogadro’s number.  Calculations using balanced chemical equations: for example, for a given mass of a reactant, calculate the volume of product.  Limiting reactants: calculate the mass of product formed when given the mass of each reactant.

Limiting Reagents Write the balanced chemical equation Write all quantitative values under equation (include units) Convert ALL reactants to moles (this is how much you HAVE) Do a mole-to-mole ratio using both reactants (this is how much you NEED) Compare HAVE to NEED to identify the limiting reactant!

Time to Practice! Grab your calculator and periodic table.

How many grams of silver can form when 12.0 g of copper are allowed to react with 15.0 g of silver nitrate? 1 Cu + 2 AgNO 3  2 Ag + 1 Cu(NO 3 ) g? g 15.0 g HAVE:

How many grams of silver can form when 12.0 g of copper are allowed to react with 15.0 g of silver nitrate? 1 Cu + 2 AgNO 3  2 Ag + 1 Cu(NO 3 ) mol ? g mol HAVE NEED: Have > Need, so L.R. = AgNO 3 Use L.R. you HAVE to calculate product that can form

Time to Practice… extra challenging! Grab your calculator and periodic table.

Limiting Reactants The reaction between solid white phosphorus (P 4 ) and oxygen gas produces solid tetraphosphorous decoxide (P 4 O 10 ). a.Determine the mass of tetraphosphorous decoxide formed if 25.0 g of phosphorous (P 4 ) and 50.0 g of oxygen are combined. b.How much of the excess reactant remains after the reaction stops?

How do we solve this? 1.What do we know? mass of phosphorous = 25.0 g P 4 mass of oxygen = 50.0 g O 2 2.Write a balanced equation for the reaction 1 P 4 (s) + 5O 2 (g)  1 P 4 O 10 (s) 25.0 g50.0 g a) ? g b) ? excess

Determine the number of moles of reactants available 1 P 4 (s) + 5O 2 (g)  1 P 4 O 10 (s) 25.0 g P 4 x 1 mol P 4 = mol P g P 4 HAVE 50.0 g O 2 x 1 mol O 2 = 1.56 mol O g O 2 HAVE 25.0 g50.0 g

1 P 4 (s) + 5 O 2 (g) ---  1 P 4 O 10 (s) Calculate the moles of O 2 needed to react completely with the available moles of P moles P 4 x 5 moles O 2 = 1.01 mol O 2 11 mol P 4 NEED

Compare the moles O 2 needed to the moles O 2 available: Need:Have: 1.01 mol O 2 <1.56 mol O 2 Because more moles of O 2 are available than are needed, the O 2 is in excess and P 4 is the limiting reactant.

Use the limiting reactant’s moles to determine how much product will be produced mol P 4 x 1 mol P 4 O 10 = mol P 4 O mol P 4 To find the mass: mol P 4 O 10 x g P 4 O 10 = 57.3 g P 4 O mol P 4 O 10

b. How much of the excess reactant remains after the reaction goes to completion? ** Difference between HAVE & NEED ** HAVE = 50.0 g O 2 & NEED = 1.01 mol Convert so units match! 50.0 g O 2 (have) – g O 2 (need) 17.7 g O 2 in excess