Balanced equations. HIGHER GRADE CHEMISTRY CALCULATIONS Calculation from a balanced equation A balanced equation shows the number of moles of each reactant.

Slides:

Advertisements

Similar presentations

MM= 342 amu Calculate the number of moles..

Advertisements

Calculations from Balanced Equations

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

Module 5.04 Gas Stoichiometry.

HONORS CHEMISTRY Feb 27, Brain Teaser Cu + 2 AgNO 3  2 Ag + Cu(NO 3 ) 2 – How many moles of silver are produced when 25 grams of silver nitrate.

Question 1: 20cm 3 of hydrochloric acid with concentration 0.5 mol/dm 3 is needed to neutralise 25 cm 3 of sodium hydroxide. What is the concentration.

CHEMISTRY February 13, 2012.

Stoichiometry Chemistry Ms. Piela.

Limiting Reactant.  Determine which reactant is left over in a reaction.  Identify the limiting reactant and calculate the mass of the product.  Calculate.

9.3 Notes Limiting reagents.

Unit 3: Chemical Equations and Stoichiometry

HIGHER GRADE CHEMISTRY CALCULATIONS Molar Volume. The molar volume is the volume occupied by one mole of a gas. Worked example 1. In an experiment the.

Chapter 12: Chemical Kinetics

Percentage yield Perform calculations to determine the percentage yield of a reaction Atom Economy Perform calculations to determine the Atom Economy of.

Calculate the Molar Mass  Magnesium chloride Ammonium sulfate  Answers: g/mol; g/mol.

STANDARD GRADE CHEMISTRY CALCULATIONS Calculations involving the mole. 1 mole of a solid substance is the formula mass of the substance in grams. This.

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.

Enthalpy of Neutralisation

Concentration of Solutions

What quantities are conserved in chemical reactions? grams and atoms.

Stoichiometry II. Solve stoichiometric problems involving moles, mass, and volume, given a balanced chemical reaction. Include: heat of reaction Additional.

Moles and reactions At the end of this section, you should be able to deduce the quantities of reactants and products from balanced equations.

Stoichiometry NC Essential Standard 2.2.4

Stoichiometry. The MOLE RATIO In a balanced chemical reaction, the coefficients tell you how many moles of each substance you need for the reaction and.

Chemistry: An Introduction to General, Organic, and Biological Chemistry, Eleventh Edition Copyright © 2012 by Pearson Education, Inc. Chapter 8 Acids.

Leave space between each step to add more information. 1.Write a balance chemical equation between the acid and the base. Remember it’s a double replacement.

Test Review Chemical Reactions and Stoichiometry.

Mole, gas volume and reactions, Chemical energy and Enthalpy,

Calculation of excess In an excess calculation you will be given

1.1.2 Moles and equations This Powerpoint contains the questions and answers for the activities 1-20.

NATIONAL 4/5 CHEMISTRY CHEMICAL CHANGES AND STRUCTURE LESSON 2 MEASURING AND CALCULATING THE RATE OF REACTION.

Chapter 1: Rate of Reaction Rate of Reaction. Which reaction is faster?

Practice questions Module 1. Basic skills M r and moles to grams Grams and moles to M r M r and grams to moles Concentration and volume to moles Concentration.

Stoichiometry. What Is It? Branch of chemistry that shows the relationships among reactants and products in a chemical reaction Equations must be balanced.

Gas Stoichiometry. We have looked at stoichiometry: 1) using masses & molar masses, & 2) concentrations. We can use stoichiometry for gas reactions. As.

EXPERIMENTAL YEILD IN THE REACTION Na 2 CO 3(aq) + CaCl 2(aq)  CaCO 3(aq) + 2 NaCl(s) OBJECTIVE : MASS OF CALCIUM CARBONATE PRODUCED DATA…THAT DAY IN.

Relative Formula Mass expressed in grams

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.

Chap. 9: Stoichiometry Identify the mole ratio of any two species in a chemical reaction. Calculate theoretical yields from chemical equations.

Remembering: Molarity and Stoichiometry Because we know you brain is getting full!!!

Stoichiometry A chemical equation shows the reactants (left side) and products (right side) in a chemical reaction. A balanced equation shows, in terms.

CHAPTER 9 Design: Winter Colors: Elemental STOICHIOMETRY.

To Do… Electronic homework (Lon-Capa) HW4 Type 1 due Monday, March 10 by 7 pm; HW4 Type 2 due Wednesday, March 12 by 7 pm HW5 Type 1 due Monday, March.

After completing this lesson you should be able to : Balanced equations show the mole ratio(s) of reactants and products. The molar volume is the same.

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.

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.

Follow the method Let the units guide you. When nitrogen and hydrogen react, they form ammonia gas, NH 3. If 56.0 g of nitrogen are used up in the reaction,

Mass-Mass Stoichiometry If the mass of any reactant or product is known for a chemical reaction, it is possible to calculate the mass of the other reactants.

Lab 8 Sodium Carbonate or Sodium Bicarbonate? Objective To determine a compound to be either Na 2 CO 3 or NaHCO 3.

Solution Stoichiometry In solution stoichiometry, we are given a concentration and a volume which we use to determine moles. n = C x V Then we use molar.

STOICHIOMETRY PRACTICE. BELLWORK #3 3/2/2011 Define the following terms: a) limiting reactant b) excess reactants c) Percent yield d) Theoretical yield.

Good Morning! Today is Tuesday, January 5, 2016 HW Due: Balancing Review Please put in the Inbox Do-Now: What do you think the term stoichiometry means?

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,

Acid–Base Reaction (Double Replacement) By Noah Carsten & Jennie Warrell.

THIS IS With Host... Your VocabularyMole to mole Limiting Reactants Mass to mass Mass to Mole Percent Yield.

Titration Calculations Revision. titration - accurate neutralisation of an acid with an alkali data obtained can be used to do calculations equation used.

Moles Noadswood Science, 2016.

Stoichiometry II.

Stoichiometry Moles to moles.

Monitoring the Rate of a Reaction

Chapter 12 Review.

Chapter 12 Review.

Solution Concentration

HIGHER GRADE CHEMISTRY CALCULATIONS

Unit 4: Chemical Equations and Stoichiometry

Unit 4: Chemical Equations and Stoichiometry

3.5 Reaction Progress Progress of a Reaction

Solution Concentration

Reacting Masses and Volumes

Presentation transcript:

Balanced equations

HIGHER GRADE CHEMISTRY CALCULATIONS Calculation from a balanced equation A balanced equation shows the number of moles of each reactant and product in the reaction. Worked example 1. The equation below shows the reaction between calcium carbonate and hydrochloric acid. CaCO 3 (s) + 2HCl(aq)  CaCl 2 (aq) + CO 2 (g) + H 2 O(l) 20g of calcium carbonate reacts with excess hydrochloric acid. Calculate (a) the mass of calcium chloride formed. (b) the volume of carbon dioxide gas formed. (Take the molar volume to be 23.0 l mol -1 ) CaCO 3 (s) + 2HCl(aq)  CaCl 2 (aq) + CO 2 (g) + H 2 O(l) Write the balanced equation Show mole ratio 1 mol  1 mol 1mol Change moles into required units 100 g  111 g 23.0 litres Use proportion 20 g  20 / 100 x111 g 20 / 100 x 23.0 litres = 22.2 g 4.6 litres

Higher Grade Chemistry Calculations for you to try. 1. The equation below shows the reaction between sodium carbonate and hydrochloric acid. Na 2 CO 3 (s) + 2HCl(aq)  2NaCl (aq) + CO 2 (g) + H 2 O(l) 2.65g of sodium carbonate reacts with excess hydrochloric acid. Calculate (a) the mass of sodium chloride formed. (b) the volume of carbon dioxide gas formed. (Take the molar volume to be 22.4 l mol -1 ) Na 2 CO 3 (s) + 2HCl(aq)  2NaCl (aq) + CO 2 (g) + H 2 O(l) 1 mol  2 mol 1mol 106 g  117 g 22.4litres 2.65 g  2.65 / 106 x117 g 2.65 / 106 x 22.4 litres = g 0.56 litres

Higher Grade Chemistry Calculations for you to try. 2.Excess sodium hydrogencarbonate is added to 200cm 3 of 0.5 mol l -1 hydrochloric acid. NaHCO 3 + HCl  NaCl + CO 2 + H 2 O Calculate the (a) mass of sodium chloride formed. (b) number of moles of water formed. (c) volume of carbon dioxide formed. (Take the molar volume of a gas to be 24 litres per mole) NaHCO 3 + HCl  NaCl + CO 2 + H 2 O 1 mol  1 mol 1mol 1 mol 1 mol  58.5 g 24 litres 1 mol The number of moles of HCl used = C x V(l) = 0.5 x 0.2 = mol  0.1 x 58.5 g 0.1 x 2.4 litres 0.1 x 1 mol = 5.85 g 2.4 litres 0.1 mol