3PbO2 + Cr2(SO4)3 + K2SO4 + H2O  3PbSO4 + K2Cr2O7 + H2SO4

Slides:



Advertisements
Similar presentations
Calculations with Equations
Advertisements

AGVISE Laboratories %Zone or Grid Samples – Northwood laboratory
Chapter 7- Molecular Formulas
Percent Yield and Limiting Reactants
Chapter 9 Chemical Quantities.
Chemical Quantities or
HOLD UP YOUR BOARD! Chapter 7 Review game.
Acid-Base Equilibria and Solubility Equilibria
The Joy of Balancing Equations. What we already know… Coefficient 2CaCl 2 The 2 tells you that there are 2Ca and 4Cl CaCl 2 Subscript The little two tells.
Section 12.2: Chemical Calculations
Chapter 7 Chemical Quantities
Solutions in Chemical Reactions
Fill in missing numbers or operations
CHAPTER 9 Water and Solutions 9.3 Properties of Solutions.
What is the % composition of carbon in benzene (C6H6)?
Chemical Quantities The Mole.
Chapter 4: Aqueous Reactions and Solution Stoichiometry
EuroCondens SGB E.
Worksheets.
Stoichiometry.
Solution Concentration Read 281 – 283. Try questions 1 – 8 (show work) Concentration = quantity of solute quantity of solution (not solvent) There are.
Multiplication X 1 1 x 1 = 1 2 x 1 = 2 3 x 1 = 3 4 x 1 = 4 5 x 1 = 5 6 x 1 = 6 7 x 1 = 7 8 x 1 = 8 9 x 1 = 9 10 x 1 = x 1 = x 1 = 12 X 2 1.
Division ÷ 1 1 ÷ 1 = 1 2 ÷ 1 = 2 3 ÷ 1 = 3 4 ÷ 1 = 4 5 ÷ 1 = 5 6 ÷ 1 = 6 7 ÷ 1 = 7 8 ÷ 1 = 8 9 ÷ 1 = 9 10 ÷ 1 = ÷ 1 = ÷ 1 = 12 ÷ 2 2 ÷ 2 =
HKCEE Chemistry Volumetric Analysis &
IIIIIIIV Ch. 10 – The Mole I. Molar Conversions A. What is the Mole? n A counting number (like a dozen) n Avogadros number (N A ) n 1 mole =
Chemical Quantities or
CALENDAR.
Amphoteric Solutions A chemical compound able to react with both an acid or a base is amphoteric.    Water is amphoteric. The two acid-base couples of.
Aqueous solutions Types of reactions
1 1  1 =.
Unit 6 Chapter 12 Chemical Quantities or
Solutions Solute – what is dissolved
Chemical Stoichiometry
Chapter 14 Section 14.2 Solution Concentration
Stoichiometry The Mole: Review A counting unit A counting unit Similar to a dozen, except instead of 12, its 602,000,000,000,000,000,000,000 Similar.
Stoichiometry.
The basics for simulations
Look at This PowerPoint for help on you times tables
Unit 1 Enthalpy.
Gas Stoichiometry Molar Volume of an Ideal Gas
Entry Task: April 27th Friday
8.4 Percent Concentration
The Mole and Stoichiometry
% Composition, Empirical and Molecular Formulas
Calculations from Balanced Equations
Mole, gas volume and reactions, Chemical energy and Enthalpy,
Excess Problems, solving for pH Chem 12 Chapter 15, Page
MaK_Full ahead loaded 1 Alarm Page Directory (F11)
Unit 11 Stoichiometry CP Chemistry.
Before Between After.
Stoichiometry.
7.85 kg KCl are dissolved in 2.38 L of solution. Find molality g table sugar (i.e., sucrose, C 12 H 22 O 11 ) are mixed into 450 g water. Find molality.
Static Equilibrium; Elasticity and Fracture
Chapter 12 Solutions Copyright © 2008 by Pearson Education, Inc.
Resistência dos Materiais, 5ª ed.
Please pick up the materials in the front and take a seat
Schutzvermerk nach DIN 34 beachten 05/04/15 Seite 1 Training EPAM and CANopen Basic Solution: Password * * Level 1 Level 2 * Level 3 Password2 IP-Adr.
1 CHAPTER THREE CHEMICAL EQUATIONS & REACTION STOICHIOMETRY CHEMICAL EQUATIONS & REACTION STOICHIOMETRY.
MOLARITY A measurement of the concentration of a solution Molarity (M) is equal to the moles of solute (n) per liter of solution M = n / V = mol / L 2.
Balanced equations. HIGHER GRADE CHEMISTRY CALCULATIONS Calculation from a balanced equation A balanced equation shows the number of moles of each reactant.
Calculate the final molarity if 75.0 mL of a 5.0 M NaOH solution were diluted to a final volume of mL. M 1 V 1 = M 2 V M NaOH.
A measurement of the concentration of a solution
1 Molarity. 2 Review of Converstions Fractions that = 1 Different units and numbers that equal each other Example: 1 in. = 2.54 cm Factors: 1 in. OR 2.54.
Molarity. Molarity (M) A concentration that expresses the moles of solute in 1 L of solution Molarity (M) = moles of solute 1 liter solution.
Basic Chemistry Copyright © 2011 Pearson Education, Inc. 1 Chapter 12 Solutions 12.6 Solutions in Chemical Reactions When a BaCl 2 solution is added to.
Steps for solving Stoichiometric Problems Involving Solution
Remembering: Molarity and Stoichiometry Because we know you brain is getting full!!!
Molarity and Stoichiometry. 1.Calculate the number of grams of sodium carbonate that are required to react fully with mL of M HCl. Na 2 CO.
If 36.2 mL of M CaCl 2 solution is added to 37.5 mL of M Na 2 CO 3, what mass of calcium carbonate, CaCO 3, will be precipitate? CaCl 2 (aq)
Presentation transcript:

3PbO2 + Cr2(SO4)3 + K2SO4 + H2O  3PbSO4 + K2Cr2O7 + H2SO4 a 34. If 25.0 g of each reactant were used in performing the following reaction, which would be the limiting reactant? (a) PbO2 (b) H2O (c) K2SO4 (d) PbSO4 (e) Cr2(SO4)3

(a) MnO2 (b) KOH (c) O2 (d) Cl2 (e) KMnO4 2MnO2 + 4KOH + O2 + Cl2  2KMnO4 + 2KCl + 2H2O b 35. If 20.0 g of each reactant were used in performing the following reaction, which would be the limiting reactant? (a) MnO2 (b) KOH (c) O2 (d) Cl2 (e) KMnO4

Which would be the limiting reactant? N2 + 3H2  2NH3 3PbO2 + Cr2(SO4)3 + K2SO4 + H2O  3PbSO4 + K2Cr2O7 + H2SO4 2MnO2 + 4KOH + O2 + Cl2  2KMnO4 + 2KCl + 2H2O CH4 + 2O2  CO2 + 2H2O

H2SO4 + 2NaOH  Na2SO4 + 2H2O EXAMPLE 3-24, P.111. Find the volume in liters and in milliliters of a 0.505 M NaOH solution required to react with 40.0 mL of 0.505 M H2SO4 solution. Answer: 80 mL NaOH Use on your own risk!

CaCO3 + 2HCl  CaCl2 + CO2 + H2O b 71. What mass of calcium carbonate, CaCO3, is required to react with 100. mL of 2.00 M HCl solution? (a) 5.00 g (b) 10.0 g (c) 15.0 g (d) 20.0 g (e) 23.0 g

2HCl + Ba(OH)2  BaCl2 + 2H2O a 72. What volume of 0.130 M HCl solution will just react with 0.424 gram of Ba(OH)2? (a) 38.1 mL (b) 32.6 mL (c) 24.1 mL (d) 18.6 mL (e) 96.7 mL

(a) 0.50 M (b) 0.10 M (c) 0.20 M (d) 0.40 M (e) 0.050 M Na2CO3 + H2SO4  Na2SO4 + CO2 + H2O e 73. If 40.0 mL of H2SO4 solution reacts with 0.212 gram of Na2CO3, what is the molarity of the H2SO4 solution? (a) 0.50 M (b) 0.10 M (c) 0.20 M (d) 0.40 M (e) 0.050 M

(a) 2.55 M (b) 6.84 M (c) 8.62 M (d) 9.78 M (e) 11.84 M a 59. Calculate the molarity of a solution that contains 70.0 g of H2SO4 in 280. mL of solution.

(a) 75.0 M (b) 0.043 M (c) 33.1 M (d) 0.067 M (e) 0.0750 M e 67. Calculate the molarity of the resulting solution if enough water is added to 50.0 mL of 4.20 M NaCl solution to make a solution with a volume of 2.80 L. Chemistry is fun!