To understand the information given in a balanced equation Objectives To understand the information given in a balanced equation To use a balanced equation to determine relationships between moles of reactant and products (mole ratio) (Stoichiometry)

Let’s create a sandwich recipe…PHeT Sandwich Shoppe! Let’s create a sandwich recipe…PHeT Slice, dozen, 100, 6.02 X 1023, one mole (pg 281)

Quinoa Side Dish www.allrecipes.com

1. Information Given by Chemical Equations The coefficients of a balanced equation give the relative numbers of molecules. A balanced chemical equation gives relative numbers (or moles) of reactant and product molecules that participate in a chemical reaction.

2. Mole-mole Relationships A balanced equation can predict the moles of product that a given number of moles of reactants will yield. How many moles 02 from __ moles H2O?

2. Mole-mole Relationships The mole ratio allows us to convert from moles of one substance in a balanced equation to moles of a second substance in the equation.

2. Mole-mole Relationships The mole ratio is determined by: Mole ratio = new mole substance A balanced mole substance A How many mole ___ from ___ moles ___? 2H2O  2H2 + O2 mole mole ratio mole

2. Mole-mole Relationships Must first have a BALANCED equation __C3H8(g) + __O2  __CO2(g) + __H2O(g) + heat Calculate the number of moles of CO2 formed when 4.30 mol C3H8 combust. mole mole ratio mole

2. Mole-mole Relationships From this new mole value, we can calculate the new recipe! C3H8(g) + 5O2  3CO2(g) + 4H2O(g) + heat 4.30 C3H8 +__O212.9 CO2 + __ H2O + heat Calculate the number of moles of H2O formed when 4.30 mol C3H8 combust. mole mole ratio mole

2. Mole-mole Relationships For all reactants AND products! C3H8(g) + 5O2  3CO2(g) + 4H2O(g) + heat 4.30 C3H8 +__O212.9 CO2 + 17.2 H2O + heat Calculate the number of moles of O2 consumed when 4.30 mol C3H8 combust. mole mole ratio mole

2. Mole-mole Relationships C3H8(g) + 5O2  3CO2(g) + 4H2O(g) + heat Calculate the number of moles of all reactants and products when 2.70 mol C3H8 combust. 2.7 C3H8 +__O2 __ CO2 + __ H2O + heat 2.7 C3H8 +13.5 O2 8.1 CO2 + 10.8 H2O + heat

2. Mole-mole Relationships Balance the following equation and determine how much NH3 can be made from 1.3 mol H2. __N2 + __H2  __NH3 N2 + 3H2  2NH3 How much N2 will be needed to completely react the 1.3 mol H2?

2. Mole-mole Relationships Using the balanced equation, determine the new mole recipe given the new mole value. 2 K + 2 H2O  H2 + 2 KOH ___K + 1.7 H2O  ___ H2 + ___KOH 1.7 K + 1.7 H2O  0.85 H2 + 1.7 KOH 4 NH3 + 5 O2  4 NO + 6 H2O ___ NH3 + ___ O2  0.6 NO + ___ H2O 0.6 NH3 + 0.75 O2  0.6 NO + 0.9 H2O

To understand the information given in a balanced equation Objectives Review To understand the information given in a balanced equation To use a balanced equation to determine relationships between moles of reactant and products (mole ratio) (Stoichiometry) Work Session: Page 287 #2-5 QUIZ!!

To perform mass calculations that involve scientific notation Objectives To learn to use Stoichiometry to relate masses of reactants and products in a chemical reaction To perform mass calculations that involve scientific notation Calculate the theoretical predications for a reaction that will be performed in the lab

Mole ratio = new mole substance A balanced mole substance A 1. Mass Calculations Stoichiometry is the process of using a balanced chemical equation to determine the relative masses of reactants and products involved in a reaction. Scientific notation can be used for the masses of any substance in a chemical equation. Mole ratio = new mole substance A balanced mole substance A

grams grams mole mole ratio mole N2 + 3H2  2NH3 1. Mass Calculations N2 + 3H2  2NH3 How much NH3 will be produced when 2.6 g H2 completely reacts with N2? **Can’t do a mass ratio!!** Must first convert from grams to moles. Don’t forget the mole! grams grams mole mole ratio mole

2.6 grams H2 grams mole mole ratio mole N2 + 3H2  2NH3 1. Mass Calculations N2 + 3H2  2NH3 How much NH3 will be produced when 2.6 g H2 completely reacts with N2? 2.6 grams H2 grams mole mole ratio mole 0.867 mol NH3: 14.71 g NH3

__ N2 + 1.3 H2  0.867 NH3 (new mol recipe) 1. Mass Calculations N2 + 3H2  2NH3 __ N2 + 1.3 H2  0.867 NH3 (new mol recipe) __ g N2 2.6 g H2 14.71 g NH3 (new g recipe) How much N2 will be consumed when 2.6 g H2 completely reacts? Use the same mole ratio— 0.433 mol N2: 12.12 g N2

0.433 N2 + 1.3 H2  0.867 NH3 (new mol recipe) 1. Mass Calculations N2 + 3H2  2NH3 0.433 N2 + 1.3 H2  0.867 NH3 (new mol recipe) 12.12 g N2 2.6 g H2 14.71 g NH3 (new g recipe) How does this relate to the Law of Conservation of Mass? Does the mass reactants = mass products? 12.12 g + 2.6 g = 14.72 g Compared to 14.71 g Why the 0.01 g difference?

1. Mass Calculations

__Al(S) + __I2(g)  __AlI3(s) 1. Mass Calculations __Al(S) + __I2(g)  __AlI3(s) Balance the equation and calculate the mass of I2 needed to completely react with 35.0 g Al. Also calculate the mass of AlI3 that will be formed. grams grams mole mole ratio mole (next)

__Al(S) + __I2(g)  __AlI3(s) (new mole) 35.0 g Balance the equation and calculate the mass of I2 needed to completely react with 35.0 g Al. Also calculate the mass of AlI3 that will be formed. 35.0 g Al (all new moles)

1.3 Al(S) + 1.95 I2(g)  1.3 AlI3(s) (new mole) 35.0 g Balance the equation and calculate the mass of I2 needed to completely react with 35.0 g Al. Also calculate the mass of AlI3 that will be formed. Now convert mole  grams for each 1.95 mol I2 = 1.3 mol AlI3 =

1.3 Al(S) + 1.95 I2(g)  1.3 AlI3(s) (new mole) 35.0 g 495.3 g 530.4 g Balance the equation and calculate the mass of I2 needed to completely react with 35.0 g Al. Also calculate the mass of AlI3 that will be formed. Check the Law of Conservation of Mass!

2. Scientific Notation (CO2 scrubber on space vehicles) __LiOH(s) + __CO2(g)  __Li2CO3(s) + __H2O(l) What mass of CO2 can 1 X 103 g LiOH absorb? What mass of H2O will be available for use? What mass of Li2CO3 will be generated? First Balance! Then, below!! grams grams mole mole ratio mole

2. Scientific Notation (CO2 scrubber on space vehicles) 2 LiOH(s) + CO2(g)  Li2CO3(s) + H2O(l) __LiOH(s) + __CO2(g)  __Li2CO3(s) + __H2O(l) What mass of CO2 can 1 X 103 g LiOH absorb? Do new mole recipe next… grams grams mole mole ratio mole

2. Scientific Notation (CO2 scrubber on space vehicles) 2 LiOH(s) + CO2(g)  Li2CO3(s) + H2O(l) 41.67LiOH + 20.8CO2 20.8Li2CO3 + 20.8H2O Now convert all moles to grams. 20.8 mol CO2 = 20.8 mol Li2CO3 = 20.8 mol H2O =

2. Scientific Notation (CO2 scrubber on space vehicles) 2 LiOH(s) + CO2(g)  Li2CO3(s) + H2O(l) 41.67LiOH + 20.8CO2 20.8Li2CO3 + 20.8H2O 1 X 103 g 915.2 g 1539.2 g 374.4 g What mass of CO2 can 1 X 103 g LiOH absorb? What mass of H2O will be available for use? What are some possible uses for the H2O? What mass of Li2CO3 will be generated? What will be done with the Li2CO3? Conserved?

2. Mass Calculations Using Scientific Notation Hydrofluoric acid, an aqueous solution containing dissolved hydrogen fluoride, is used to etch glass by reacting with silica, SiO2, in the glass to produce gaseous silicon tetrafluoride and liquid water. The unbalanced reaction is: __HF(aq) + __SiO2(s)  __SiF4(g) + __H2O(l) Calculate the mass of HF that is needed and the products that are produced to react completely with 5.68 g SiO2.

2. Mass Calculations Using Scientific Notation 4 HF(aq) + SiO2(s)  SiF4(g) + 2 H2O(l) __ HF(aq) + __ SiO2(s)  __ SiF4(g) + __ H2O(l) 5.68 g SiO2 = Mole ratio grams grams mole mole ratio mole

2. Mass Calculations Using Scientific Notation 4 HF(aq) + SiO2(s)  SiF4(g) + 2 H2O(l) 0.378 9.45 X10-2  9.45 X10-2 + 0.189 (moles) 7.56 g 5.68 g  9.828 g 3.41 g (grams) Was mass Conserved? grams grams mole mole ratio mole

2. Mass Calculations Comparisons Which is the better antacid? Baking Soda,NaHCO3,or MOM, Mg(OH)2? How many moles of HCl will react with 1.00 g of each antacid? NaHCO3(s) + HCl(aq)  NaCl(aq) + H2O(l) + CO2(g) Demo Mg(OH)2(s) + 2HCl(aq)  2H2O(l) + MgCl2(aq) grams grams mole mole ratio mole

2. Mass Calculations Comparisons Which is the better antacid? Baking Soda,NaHCO3,or MOM, Mg(OH)2? How many moles of HCl will react with 1.00 g of each antacid? NaHCO3(s) + HCl(aq)  NaCl(aq) + H2O(l) + CO2(g) 1.0 g/1.19 X 10-2 mol NaHCO3 1.19 X 10-2 mol HCl Mg(OH)2(s) + 2HCl(aq)  2H2O(l) + MgCl2(aq) 1.0 g/1.71 X 10-2 mol Mg(OH)2 3.42 X 10-2 mol HCl MOM is better!

2. Mass Calculations for Lab- don’t round too much! Predict the charged balanced products and their solubility. Then balance the equation, and calculate the number of grams of all reactants and products needed to completely react with 0.01 mol SrCl2. ___SrCl2 + ___Na2CO3

2. Mass Calculations for lab SrCl2 + Na2CO3 SrCO3 + 2NaCl 0.01 SrCl2 + ___Na2CO3___SrCO3+___NaCl 0.01SrCl2 + 0.01Na2CO3  0.01SrCO3 + 0.02NaCl 1.59 g + 1.06 g  1.48 g + 1.17 g Conservation of mass? Theoretical vs Real…

2. Mass Calculations for lab Calculate the amount of grams contained in 0.01 mol the following hydrated crystals: SrCl2 x 6H2O Na2CO3 X H2O

2. Mass Calculations for lab 0.01SrCl2 x 6H2O + 0.01Na2CO3 X H2O  0.01SrCO3 + 0.02NaCl 2.67 g + 1.24 g  1.48 g + 1.17 g

To understand the information given in a balanced equation Objectives Review To understand the information given in a balanced equation To use a balanced equation to determine relationships between moles of reactant and products (mole ratio) (Stoichiometry) Calculate the theoretical predications for a reaction that will be performed in the lab Work Session: Page 295 1, 2, 4, 5, 6

To understand the concept of limiting reactants Objectives To understand the concept of limiting reactants To learn to recognize the limiting reactant in a reaction To learn to use the limiting reactant to do stoichiometric calculations To learn to calculate percent yield

1. The Concept of Limiting Reactants Stoichiometric mixture (balanced) N2(g) + 3H2(g)  2NH3(g)

1. The Concept of Limiting Reactants Limiting reactant mixture (runs out first) N2(g) + 3H2(g)  2NH3(g)

Mixture of CH4 and H2O Molecules Reacting CH4 + H2O  3H2 + CO

CH4 + H2O  3H2 + CO

The amount of products that can form is limited by the methane. Limiting Reactants The amount of products that can form is limited by the methane. Methane is the limiting reactant. Water is in excess. Limiting and Excess Reactants depend on the actual amounts of reactants present and must be calculated for each different reaction. (Water won’t always be in excess!)

Limiting Reactants

2. Calculations Involving a Limiting Reactant What mass of water is needed to react with 249 g methane (CH4)? CH4 + H2O  3H2 + CO 249 g CH4 How many g of H2 and CO are produced? 279 g H2O, 93 g H2, 434 g CO

2. Calculations Involving a Limiting Reactant How many g of H2 and CO are produced when 249 g methane (CH4) reacts with 300 g H2O? CH4 + H2O  3H2 + CO 249 g + 279 g  93 g + 434 g (last slide) 249 g + 300 g  93 g + 434 g The 249 g CH4 will react with only 279 g H2O, thereby leaving 21 g H2O in excess. CH4 is the reactant that will run out first and is the LIMITING REACTANT.

2. Calculations Involving a Limiting Reactant So, how do we figure this out using Stoich? 1) Balance Reaction 2) Do Stoich to relate EACH reactant to one product 3) Whichever reactant produces the LEAST amount of product is the LIMITING REACTANT

2. Calculations Involving a Limiting Reactant

2. Calculations Involving a Limiting Reactant Suppose 2.50 X 104 g N2 reacts with 5 X 103 g H2. Determine the limiting reactant. ___N2 + ___H2  ___NH3 2.50 X 104 g N2 5 X 103 g H2 From N2  1785.72 mol NH3 From H2  1666.67 mol NH3 ; 28333.3 g NH3 Not just because original mass of H2 was less.

2. Calculations Involving a Limiting Reactant How much N2 will be produced when 18.1 g NH3 and 90.4 g CuO react? Determine the limiting reactant. __NH3 + __CuO  __N2 + __Cu + __H2O 18.1 g NH3 90.4 g CuO NH3 0.53 mol N2 CuO 0.38 mol N2 *limiting reactant* 10.64 g N2

3. Percent Yield Theoretical Yield The maximum amount of a given product that can be formed when the limiting reactant is completely consumed. Just what we’ve been calculating! The actual yield (amount produced is usually given) of a reaction is usually less than the max theoretical yield because of side or incomplete reactions. Percent Yield The actual amount of a given product as the percentage of the theoretical yield.

3. Percent Yield From the previous set of calculations, we determined that 10.64 g of N2 would theoretically be produced from the reaction: 2NH3 + 3CuO  N2 + 3Cu + 3H2O What would the Percent Yield be if 7.25 g of N2 were actually produced? 7.25 g N2 X 100 = 68.14 % 10.64 g N2

Consider the following reaction: __CO(g) + __H2(g)  __CH3OH(l) 3. Percent Yield Consider the following reaction: __CO(g) + __H2(g)  __CH3OH(l) If 6.85 X 104 g CO reacts with 8.6 X 103 g H2 to produce an actual yield of 3.57 X 104 g CH3OH, determine the limiting reactant, the theoretical yield, and the percent yield.

2446.4 mol CH3OH 2150 mol CH3OH *Limiting 3. Percent Yield CO(g) + 2H2(g)  CH3OH(l) 2446.4 mol CO 4300 mol H2 2446.4 mol CH3OH 2150 mol CH3OH *Limiting 2150 mol CH3OH *32g/mol = 68800 g CH3OH % yield = 3.57 X 104 g/ 6.88 X 104 g = 51.89 % 28 g/mol CO: 2 g/mol H2: 32 g/mol CH3OH

Consider the following reaction: __TiCl4 + __O2  __TiO2 + __Cl2 3. Percent Yield Consider the following reaction: __TiCl4 + __O2  __TiO2 + __Cl2 If 6.71 X 103 g TiCl4 reacts with 2.45 X 103 g O2 with a percent yield of 75%, determine the limiting reactant, the theoretical yield, and the actual yield of TiO2. TiCl4 is limiting 2119.6 g TiO2 is actually produced in this reaction. 189.68 g/mol TiCl4: 32 g/mol o2: 79.88 g/mol TiO2

To understand the concept of limiting reactants Objectives Review To understand the concept of limiting reactants To learn to recognize the limiting reactant in a reaction To learn to use the limiting reactant to do stoichiometric calculations To learn to calculate percent yield Work Session: Page 308 2, 3, 4, 5