Challenge Stoichiometry

Slides:



Advertisements
Similar presentations
TIER 6 Combine the knowledge of gases and solutions to perform stoichiometric calculations.
Advertisements

Stoichiometry Problem
(STOY-KEE-AHM-EH-TREE). Stoichiometry is the part of chemistry that studies amounts of reactants and products that are involved in reactions. Chemists.
Limiting Reactant. Which will run out first? In the real world, one reactant will be consumed before other(s) In the real world, one reactant will be.
Stoichiometry TJ Bautista, Sean Higgins, Joanna Lee Period 6.
Lecture 109/21/05. Mass reactant Stoichiometric factor Moles reactant Moles product Mass product Stoichiometric Calculations.
Is this a balanced eq ? We only need ONE information during the reaction (reacted or produced)
Theoretical Yield The amount of product(s) calculated using stoichiometry problems.
Atomic Mass l Atoms are so small, it is difficult to discuss how much they weigh in grams l Use atomic mass units. l an atomic mass unit (amu) is one.
Stoichiometry Chapter 8. Stoichiometry Chemical equations Limiting reagent Problem types Percent yield Mass-mass Mole - mole other.
 All stoichiometry problems start with a balanced equation.  You must check for this!  If it is not balanced, BALANCE IT!
The substance that limits the amount of product that can be made
Solutions: Molarity. A. Concentration – measure of the amount of solute that is dissolved in a given amount of solvent I. Concentration of Solutions Solutions:
John E. McMurry Robert C. Fay C H E M I S T R Y Sixth Edition Chapter 3 Mass Relationships in Chemical Reactions These Sections Will NOT be on EXAM 1.
Calculate the mass of Cu produced? Mass of beaker and Cu – mass of beaker.
Define mole ratio (What is it? How is it determined?)
= Ω g Cu x 63.5 g 1 mole x 2 mol Cu 1 mol Cu 2 O x 1mol g Ω g Cu 2 O = Ω g Cux 63.5 g 1 mole x 1 mol Cu 1 mol CuO x 1mol 79.5.
Conversion Stoich. Moles to Quantity Moles of 1 substance are given and you are asked to find the mass, concentration or volume of another.
STOICHIOMETRY. The Problem  If 41.6 g of N 2 O 4 reacts with 20.8 g of N 2 H 4 the products will be nitrogen and water. What mass of water will be produced?
1 Pb(NO 3 ) 2 (aq) + KI (aq)  PbI 2 (s) + KNO 3 (aq) __ 212 Molarity and Stoichiometry M M V V P P mol M L M = mol L mol = M L What volume of 4.0 M KI.
Students type their answers here
Unit 8 Review Stoichiometry. 1. Describe how a chemist uses stoichiometry? To determine the amount of reactants needed or products formed based on the.
Challenge Stoichiometry Lesson 7. A 10.0 g sample of a Cu/Ag alloy reacted with concentrated HNO 3 according to the following equations. A total of
Chem 105 Chpt 4 Lsn 12 1 CHAPTER 5 Reactions in Aqueous Solution Homework for Chapter 5 problems posted due Feb 26 Know: Table 5.2, pg 182 Use: Figure.
Practice Suppose that a solution containing 3.50 grams of Na3PO4 is mixed with a solution containing 6.40 grams of Ba(NO3)2. Na3PO4 + Ba(NO3)2  Ba3(PO4)2.
Chemical Reactions Chapter 10 Section 1.
Reactions in Aqueous Solution
Practice Problems Chang, Chapter 4 Reactions in Aqueous Solutions
Calculations from Equations Chapter 9
Stoichiometry.
Electrochemistry.
MASS - MASS STOICHIOMETRY
Percent Copper in Brass
Predicting Products Chapter 11.
MOLARITY.
REDOX REACTIONS REACTIONS Day 1 Review Batteries.
Chapter 12 Stoichiometry
(STOY-KEE-AHM-EH-TREE)
Limiting and Excess Reactants
Introduction to Chemical Reactions: Copper Cycle
ICE Charts With Acid Base Reactions.
Solution Stoichiometry
Stoichiometry Chapter 12 the relationship between the relative quantities of substances taking part in a reaction or forming a compound, typically a ratio.
Solution Stoichiometry
Chemistry 11 Challenge Question
Chapter 7 Reactions in solution.
Visualizing Limiting Reactant
CHAPTER 11 Stoichiometry 11.2 Percent Yield and Concentration.
Chemical Equations Stoichiometry V
Chapter 12 Stoichiometry 12.2 Chemical Calculations
Chemical Calculations
Chapter 12 Stoichiometry 12.2 Chemical Calculations
Chapter 12 Stoichiometry 12.2 Chemical Calculations
ICE Charts With Acid Base Reactions.
Signs of a Chemical Change (Chemical Reaction)
Chapter 12 Stoichiometry 12.2 Chemical Calculations
Volumetric Analysis Using titration of a solution with known concentration to determine information about a different solution of unknown concentration.
Unit 1: Reaction Kinetics
What Really Happens in Stoichiometry, Most of the Time
Oxidation Numbers.
(7.1) Stoichiometry-Part 2
Chapter 12 Stoichiometry 12.2 Chemical Calculations
Writing and Balancing Chemical Reactions
Chapter 12: Stoichiometry
Identifying Chemical Reactions.
Molarity = mol/L Molarity = moles of solute / Liters of solution
Solution Stoichiometry
Limiting Reactants, Excess Reactants and Theoretical Yield
Chapter 11 Review.
Molarity and Stoichiometry
Presentation transcript:

Challenge Stoichiometry Lesson 7

A 10.0 g sample of a Cu/Ag alloy reacted with concentrated HNO3 according to the following equations. A total of 8.986 g of NO2 was isolated. Assuming a 100 % yield of NO2, determine the mass of Cu and Ag in the alloy. Cu(s) + 4HNO3(aq) → Cu(NO3)2(aq) + 2NO2(g) + 2H2O(l) Ω g ?g 2Ag(s) + 4HNO3(aq) → 2AgNO3(aq) + 2NO2(g) + 2H2O(l) (10 - Ω) g ?g 10.0 g 8.986 g

A 10.0 g sample of a Cu/Ag alloy reacted with concentrated HNO3 according to the following equations. A total of 8.986 g of NO2 was isolated. Assuming a 100 % yield of NO2, determine the mass of Cu and Ag in the alloy. Cu(s) + 4HNO3(aq) → Cu(NO3)2(aq) + 2NO2(g) + 2H2O(l) 2Ag(s) + 4HNO3(aq) → 2AgNO3(aq) + 2NO2(g) + 2H2O(l) Ω g Cu x 1 mole x 2 mole NO2 x 46.0 g = 1.4488 Ω 63.5 g 1 mole Cu 1 mole 10 – Ω g Ag x 1 mole x 2 mole NO2 x 46.0 g = 4.2632 - 0.42632 Ω 107.9 g 2 mole Ag 1 mole 1.4488 Ω + 4.2632 - 0.42632 Ω = 8.986 1.0225 Ω = 4.7228 Note the loss of a sig fig! Ω = 4.62 g Cu 10.0 - 4.62 = 5.4 g Ag

A 10.00 g sample of brass, a Cu/Zn alloy, is completely reacted with concentrated HNO3 according to the following equations. A total of 14.2356 g of NO2 was isolated. Assuming a 100 % yield of NO2, determine the mass of Cu and Zn in the alloy. Cu(s) + 4HNO3(aq) → Cu(NO3)2(aq) + 2NO2(g) + 2H2O(l) Ω g ?g Zn(s) + 4HNO3(aq) → Zn(NO3)2(aq) + 2NO2(g) + 2H2O(l) (10 - Ω) g ?g 10.0 g 14.2356 g

A 10.00 g sample of brass, a Cu/Zn alloy, is completely reacted with concentrated HNO3 according to the following equations. A total of 14.2356 g of NO2 was isolated. Assuming a 100 % yield of NO2, determine the mass of Cu and Zn in the alloy. Cu(s) + 4HNO3(aq) → Cu(NO3)2(aq) + 2NO2(g) + 2H2O(l) Zn(s) + 4HNO3(aq) → Zn(NO3)2(aq) + 2NO2(g) + 2H2O(l) Ω g Cu x 1 mole x 2 mole NO2 x 46.0 g = 1.4488 Ω 63.5 g 1 mole Cu 1 mole 10 – Ω g Zn x 1 mole x 2 mole NO2 x 46.0 g = 14.0673 - 1.40673 Ω 65.4 g 1 mole Zn 1 mole 1.4488 Ω + 14.0673 - 1.40673 Ω = 14.2356 0.04207 Ω = 0.1683 Ω = 4.00 g Cu 10.00 – 4.00 = 6.00 g Zn