Performing Calculations involving Limiting and Excess Reagents.

Slides:

Advertisements

Similar presentations

Limiting Reactants and Percent Yield

Advertisements

Limiting Reactants & Percent Yield

Limiting Reactants and Percent Yield

Copyright©2004 by Houghton Mifflin Company. All rights reserved. 1 Introductory Chemistry: A Foundation FIFTH EDITION by Steven S. Zumdahl University of.

Chemical Quantities Chapter 9

Mass relationships in chemical reactions: Stoichiometry

Limiting Reactants & Percent Yield. Limiting Reactants  The reactant that limits the amount of product formed in a chemical reaction. The quantity of.

Stoichiometry Introduction.

Section “Limiting” Reagent

Percent Yield and Limiting Reactants

Limiting Reactants and Percent Yields

Limiting and Excess Reactants

Limiting Reactants and Excess

Limiting Reactants & Percent Yield

Who can propel a rocket the farthest? The best scientist will win!

Limiting Reagents Excess Reagents Percent yield

Limiting Reactant and Percent Yield Limiting Reagent u If you are given one dozen loaves of bread, a gallon of mustard and three pieces of salami, how.

Chapter 9 – Review Stoichiometry

Limiting Reagents Determine the mass of P 4 O 10 formed if 25g P 4 and 50g O 2 are combined. P 4 + 5O 2  P 4 O 10 What is the given? We have to find the.

Limiting Reagent and Percent Yield

Limiting Reagent u The limiting reagent is the reactant you run out of first. u The excess reagent is the one you have left over. u The limiting reagent.

Review Answers with step-by-step examples

Lecture 129/28/05 So what have you learned in the last few days?

2Al + 3Br 2  2Al Br g 46.0g ?g 1.How much aluminum bromide can be formed from aluminum? 2.How much aluminum bromide can be formed from bromine?

Brought to you by Coach Cox PERCENT YIELD. WHAT IS PERCENT YIELD? Theoretical Yield – the maximum amount of product that can be produced from a given.

P ERCENT Y IELD. OBJECTIVE I can calculate percent yield of a reaction.

*Notice, kind of like the opposite of Percent Error.

11-3 Limiting Reactants Reactions to this point have had one ingredient in excess. When you are given two or both reactant amounts, you have a limiting.

Chapter 12 Review “Stoichiometry”

Objectives To understand the concept of limiting reactants

Percent Yield and Limited Reactants NC Essential Standard Analyze the stoichiometric relationships inherent in a chemical reaction.

Chapter 12 Stoichiometry. 1. The part of chemistry that deals with the amount of substances involved in chemical reactions A. 3 basic steps to every stoichiometry.

Theoretical yield vs. Actual yield. Suppose the theoretical yield for an experiment was calculated to be 19.5 grams, and the experiment was performed,

Reaction Stoichiometry. Objectives Understand the concept of stoichiometry. Be able to make mass-to-mass stoichiometric calculations.

Limiting reagents In lab a reaction is rarely carried out with exactly the required amounts of each reactant. In lab a reaction is rarely carried out with.

Limiting Reactants and Excess What is the Limiting Reagent (Reactant)? It is the substance in a chemical reaction that runs out first. The limiting reactant.

Chapter 9, section 3, part 2 Percent Yield. Why percent yield?  Usually, not all the product possible is actually formed.  theoretical yield  maximum.

Percent Yield. “yield—” the amount of product actually made through a chemical reaction. Why is this value important? Theoretical yield— calculated amount.

Limiting Reactants & Percent Yield. Definitions  Limiting Reactant - completely consumed in the reaction and determines the amount of product formed.

Limiting Reactants Unit 7 1. Limiting Reactants 2.

Limiting Reagents Stochiometry Chapter 9 Chemical Reactions.

Ch. 9-3 Limiting Reactants & Percent Yield. POINT > Define limiting reactant POINT > Identify which reactant is limiting in a reaction POINT > Define.

SOL Review 6 Stoichiometry. Consider: 4NH 3 + 5O 2  6H 2 O + 4NO Many conversion factors exist: 4 NH 3 6 H 2 04NO 5O 2 (and others) 5 O 2 4 NO4 NH 3.

Challenge Problem When nitrogen and hydrogen react, they form ammonia gas, which has the formula NH 3. If 56.0 g of nitrogen are used up in the reaction,

Mass-Mass Conversions 56.0 g N 2 x g N 2 g NH = 1904 = When nitrogen and hydrogen react, they form ammonia gas, which has the formula.

Unit 6: Limiting Reactants and Percent Yield Chapter 11.3 and 11.4.

WARM UP The reaction of iron ore with carbon follows the equation: 2 Fe 2 O C  4 Fe + 3 CO 2. How many moles of Fe can be produced from 37 moles.

LIMITING REACTIONS INB PAGE 43. ESSENTIAL QUESTION: Why is the limiting reactant not always the reactant with fewer moles?

Stoichiometry and the Mole

Sec 12.3 limiting reactant, percent, actual and theoretical Yield

Chemistry Limiting Reactions.

Finding the Amount of Excess Reactant Left Over

Finding the Amount of Excess Reactant Left Over

Theoretical Yield.

Section 11.3 Limiting Reactants

Stoichiometry Comes from the Greek words stoicheion, meaning “element,” and metron, meaning “measure.”

*Notice, kind of like the opposite of Percent Error.

PROBLEM 4 OF LIMITING REAGENT SET

Bell Work: Limiting Reactant Problems

Limiting Reagent 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.

2S (s) + 3O2 (g) 2SO3 (g) O < 1

Stoichiometry of Limiting and Excess Quantities

#5 of 11-6 LIMITING REAGENT SET

1.50 g. NH mol NH ( ) 1 mol = 17.00g STEP 1-Convert to moles

Cu (s) + 2AgNO3(aq) → Cu(NO3)2 (aq) + 2Ag(s)

LIMITING REAGENT (LR) gets completely used up in the reaction

2HCl + Ca(OH)2 → CaCl2 + 2H2O We don’t have enough calcium hydroxide

Reaction Stoichiometry

Presentation transcript:

Performing Calculations involving Limiting and Excess Reagents

__ Al (s) +__ O 2(g)  __Al 2 O 3(s) 1.You have a sealed vessel that contains 623 mL of oxygen and 111g aluminum. How much aluminum oxide can I produce? 2.How much of your excess reactant is left over? 3.If you collected 1.06g of aluminum oxide, what was the percent yield of the reaction? 234

1.You have a sealed vessel that contains 623 mL of oxygen and 111g aluminum. How much aluminum oxide can I produce? __ Al (s) +__ O 2(g)  __Al 2 O 3(s) 234 WHAT IS YOUR LIMITING REACTANT?? 623 mL O 2 x 1 mol O L 1 L O mL x = 111g Al x 1 mol Al 26.98g = mol O mol Al HAVE AVAILABLE x 4 mol Al 3 mol O 2 = mol Al In order to use up all the oxygen available, you would need mol Al. So you should ask yourself, it is going to be possible to use up all the O 2 ?? YES! Therefore, oxygen is limiting and aluminum is in excess.

1.You have a sealed vessel that contains 623 mL of oxygen and 111g aluminum. How much aluminum oxide can I produce? __ Al (s) +__ O 2(g)  __Al 2 O 3(s) 234 Oxygen is limiting and aluminum is in excess. The amount of product you can form is going to be dependent on the reactant that will get used up! Therefore, it is dependent on your limiting reactant mol O 2 x 2 mol Al 2 O 3 3 mol O 2 = x g Al 2 O 3 1 mol Al 2 O g Al 2 O 3 formed

2. How much of your excess reactant is left over? __ Al (s) +__ O 2(g)  __Al 2 O 3(s) 234 The amount of excess reactant left over is dependent upon how much we had to start and how much was used during the reaction mol O 2 x 4 mol Al 3 mol O 2 = mol Al We already know that oxygen in our limiting reactant. Therefore, when it is completely used in the reaction, mol of aluminum is also used. According to our original calculations, we originally had 4.11 mol of aluminum mol Al – mol Al = 3.74 mol Al remains unused 3.74 mol Al 26.98g 1 mol Al x = 101g Al remaining

3. If you collected 1.06g of aluminum oxide, what was the percent yield of the reaction? __ Al (s) +__ O 2(g)  __Al 2 O 3(s) 234 In question 1, you’ve already determined that the amount of aluminum oxide produced was 1.89g. This is your THEORETICAL YIELD Percent Yield = Actual yield Theoretical yield x 100% Percent Yield = 1.06 g 1.89g x 100% = 56.1%