Stoichiometry. Warm-Up Example of stoich problem Magnesium metal burns in the air Mg (s) + O 2(g) MgO (s) 2 Mg (s) + O 2(g) 2 MgO (s) 3g? g.

Slides:



Advertisements
Similar presentations
Limiting Reactants & Percent Yield
Advertisements

Mass relationships in chemical reactions: Stoichiometry
Stoichiometry, Limiting Reactants and Percent Yield
Stoichiometry Introduction.
WE BE COOKING…. OH CRAP I ONLY HAVE ½ CUP OF SUGAR! Limiting Reagents.
Stoichiometry.
How many moles of water will be produced when 8 grams of hydrogen gas react with the oxygen in the air? Episode 801.
Stoichiometry – “Fun With Ratios”
Limiting Reagent and Theoretical Yield 1
Stoichiometry of Chemical Equations and Formulas.
Limiting Reagent and Percent Yield
Review Answers with step-by-step examples
Theoretical Yield The amount of product(s) calculated using stoichiometry problems.
12.3 Limiting Reagent and Percent Yield
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.
Stoichiopardy Holy Moley Do the 2 or 3 step Random Limit my Percent Q $100 Q $200 Q $300 Q $400 Q $500 Q $100 Q $200 Q $300 Q $400 Q $500 Final Chempardy.
Percentage Yield.
Mass Relationships in Chemical Reactions Chapter 3.
April 7, 2014 Today: Stoichiometry and % Yield. Percent Yield Remember, stoichiometry is used to tell you how much product you can form from X amount.
Stoichiometry Chapter 9. Step 1 Balance equations and calculate Formula Mass (FM) for each reactant and product. Example: Tin (II) fluoride, SnF 2, is.
Stoichiometry – “Fun With Ratios” Main Idea: The coefficients from the balanced equation tell the ratios between reactants and products. This ratio applies.
Stoichiometric Calculations Stoichiometry – Ch. 8.
Stoichiometry. Information Given by the Chemical Equation  The coefficients in the balanced chemical equation show the molecules and mole ratio of the.
II. Stoichiometry in the Real World Stoichiometry – Ch. 11.
Gravimetric Stoichiometry Is used to calculate masses of reactants and products in a reaction.
Limiting Reactions and Percent Yield Calculating by moles or mass ©2011 University of Illinois Board of Trustees
Unit 8 Review Stoichiometry Complete on Markerboard or in your notes.
Chapter 9 Stoichiometry. 9.1 Intro. To Stoichiometry What is Stoichiometry? – The study of the quantitative relationships that exist in chemical formulas.
Stoichiometry: A calculation based on a balanced equation. Granada Hills Charter High School.
No Bell Ringer Today. We will have a test next Tuesday.
Stoichiometry Warmup I have 1 mole of CO 2 gas at STP. How many grams of CO 2 do I have? How many Liters of CO 2 do I have? How many molecules of CO 2.
Honors Stoichiometry Examples. Example 1 What mass of carbon dioxide in grams is needed to react with 3.00mol of H 2 O in the following photosynthetic.
PERCENT YIELD. Percent Yield Percent yield is the ratio of actual yield of an experiment to theoretical yield. To find percent yield, you divide the actual.
Stoichiometry Unit 8/9. Stoichiometry The process of determining how much product is made or how much reactant is needed during a chemical reaction.
Percent Yield. “yield—” the amount of product actually made through a chemical reaction. Why is this value important? Theoretical yield— calculated amount.
Limiting Reactant. Stoichiometry Refresher Recall: The mole ratio can be used to determine the quantity (moles) or mass of reactants consumed and products.
Percent Yield and Limiting Stoichiometry. EXAMPLE QUESTION #1 (Limiting Stoichiometry ICE BOX)Limiting Stoichiometry Calculate the theoretical yield of.
Percentage Yield.
Stoichiometry Chapter 9 Percent Yield Stoich ppt _5 Percent Yield.
Chemistry Chapter 9 - Stoichiometry South Lake High School Ms. Sanders.
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.
Unit 8 Review Stoichiometry. 1. Describe how a chemist uses stoichiometry? To determine the amount of reactants needed or products formed based on the.
Stoichiometry Page 39. Essential Question How can the mole ratio of a chemical equation be determined and used for conversions?
Stoichiometry in the Real World Stoichiometry – Ch. 11.
Honors Stoichiometry Examples. Example 1 What mass of carbon dioxide in grams is needed to react with 3.00mol of H 2 O in the following photosynthetic.
Section 9.2 Equation Stoichiometry
Let’s talk… mole to mole
Stoichiometry: the study of the quantitative relationships that exist between the amounts of reactants and products in a chemical reaction. Stoichiometry.
Calculations from Chemical Equations
Percent Yield in a Chemical Reaction.
Limiting Reactant.
Calculations using balanced chemical equations Revision
Percent Yield.
Unit 8: Stoichiometry: Part 1
[
STOICHIOMETRY.
Theoretical Yield & Percent Yield
Stoichiometry Comes from the Greek words stoicheion, meaning “element,” and metron, meaning “measure.”
II. Stoichiometry in the Real World (p. 368 – 375)
PROBLEM 4 OF LIMITING REAGENT SET
Unit 4 Stoichiometry Stoichiometry, mol-mol Stoich. , volume-volume
How many moles of water are made by
STOICHIOMETRY – To determine, using stoichiometric calculations, the quantity of a substance involved in a chemical reaction. 5.8 – To solve numerical.
Bellwork Tuesday 5.9 L of carbon dioxide is combined with 8.4 g MgO in a synthesis reaction to form magnesium carbonate. How many grams of magnesium carbonate.
% Yield Definitions 1. Theoretical Yield
PROBLEM 4 OF LIMITING REAGENT SET
Balancing Chemical Equations
Chemical Calculations
Experiment 3 Determining an Empirical Formula
Presentation transcript:

Stoichiometry

Warm-Up

Example of stoich problem Magnesium metal burns in the air Mg (s) + O 2(g) MgO (s) 2 Mg (s) + O 2(g) 2 MgO (s) 3g? g

Example of stoich problem 2 Mg (s) + O 2(g) 2 MgO (s) 3 g? g 3 g 1 mol 24.3 g 24.3 g/mol 40.3 g/mol mol 1 mol 40.3 g

Example of stoich problem 2 Mg (s) + O 2(g) 2 MgO (s) 3 g 4.96 g 3 g 1 mol 24.3 g 24.3 g/mol 40.3 g/mol mol 1 mol 40.3 g

Percent Yield 2 Mg (s) + O 2(g) 2 MgO (s) 3 g 4.96 g (theoretical) 24.3 g/mol 40.3 g/mol mol 3.72 g (actual) 3.72/4.96 x 100% = 75%

Questions that can be asked 2 Mg (s) + O 2(g) 2 MgO (s) 3 g? g 24.3 g/mol 40.3 g/mol mol

Questions that can be asked 2 Mg (s) + O 2(g) 2 MgO (s) 3 g? g 24.3 g/mol 32.0 g/mol mol How much oxygen is used in this reaction?

Questions that can be asked 2 Mg (s) + O 2(g) 2 MgO (s) 3 g? g 24.3 g/mol 32.0 g/mol mol 3 g Mg 1 mol Mg 24.3 g Mg 2 mol Mg 1 mol O 2 = 1.97 g O 2 1 mol O g O 2 mole ratio