*STOICHIOMETRY TUTORIAL

oleh :Trisna Kumala Dhaniswara,S.T.,M.T

*Instructions: This is a work along tutorial. Each time you click the mouse or touch the space bar on your computer, one step of the problem solving occurs. Pressing the PAGE UP key will backup the steps.

Get a pencil and paper, a periodic table and a calculator, and lets get to work.

*(1-2-3) General Approach For Problem Solving:1. Clearly identify the Goal or Goals and the UNITS involved. (What are you trying to do?)2. Determine what is given and the UNITS.3. Use conversion factors (which are really ratios) and their UNITS to CONVERT what is given into what is desired.

*Table of Contents: Click on each tab to view problem types.Sample problem 1Sample problem 2Converting grams to molesMole to Mole Conversions View Complete Slide ShowGram-Mole and Gram-Gram ProblemsSolution Stoichiometry ProblemsLimiting/Excess/ Reactant and Theoretical Yield Problems :

*Sample problem for general problem solving.Sam has entered into a 10 mile marathon. Use ALL of the following conversions (ratios) to determine how many inches there are in the race.5280 ft = 1 mile; 12 inches = 1 ft1. What is the goal and what units are needed?Goal = ______ inches2. What is given and its units?10 miles3. Convert using factors (ratios).10 miles = inches633600Units matchGivenGoalConvertMenu

*Sample problem #2 on problem solving.A car is traveling at a speed of 45 miles per hr (45 miles/hr). Determine its speed in kilometers per second using the following conversion factors (ratios). 1 mile = 5280 ft; 1 ft = 12 in; 1 inch = 2.54 cm; k = 1 x 103; c = 1 x 10-2; 1 hr =60 min; 1 min = 60 s= km s0.020Units Match!Goalc cancels cm remainsGivenThis is the same as putting k over k

*Converting grams to moles.Determine how many moles there are in 5.17 grams of Fe(C5H5)2.Goal= moles Fe(C5H5)2Given5.17 g Fe(C5H5)2Use the molar mass to convert grams to moles.Fe(C5H5)22 x 5 x 1.001 = 10.012 x 5 x 12.011 = 120.11 1 x 55.85 = 55.850.0278units match

*Stoichiometry (more working with ratios)Ratios are found within a chemical equation.2HCl + Ba(OH)2 2H2O + BaCl2 112 moles of HCl react with 1 mole of Ba(OH)2 to form 2 moles of H2O and 1 mole of BaCl2 coefficients give MOLAR RATIOS

*When N2O5 is heated, it decomposes:2N2O5(g) 4NO2(g) + O2(g)a. How many moles of NO2 can be produced from 4.3 moles of N2O5?= moles NO24.3 mol N2O58.6b. How many moles of O2 can be produced from 4.3 moles of N2O5?= mole O24.3 mol N2O52.22N2O5(g) 4NO2(g) + O2(g)4.3 mol? mol2N2O5(g) 4NO2(g) + O2(g)4.3 mol? molMole Mole ConversionsUnits match

*When N2O5 is heated, it decomposes:2N2O5(g) 4NO2(g) + O2(g)a. How many moles of N2O5 were used if 210g of NO2 were produced?= moles N2O5210 g NO22.28b. How many grams of N2O5 are needed to produce 75.0 grams of O2?= grams N2O575.0 g O2506gram mole and gram gram conversions 2N2O5(g) 4NO2(g) + O2(g)210g? moles2N2O5(g) 4NO2(g) + O2(g)75.0 g? gramsUnits match

*Aluminum is an active metal that when placed in hydrochloric acid produces hydrogen gas and aluminum chloride. How many grams of aluminum chloride can be produced when 3.45 grams of aluminum are reacted with an excess of hydrochloric acid?First write a balanced equation. Al(s) + HCl(aq) AlCl3(aq) + H2(g)2 6 2 3Gram to Gram Conversions

*Aluminum is an active metal that when placed in hydrochloric acid produces hydrogen gas and aluminum chloride. How many grams of aluminum chloride can be produced when 3.45 grams of aluminum are reacted with an excess of hydrochloric acid? Al(s) + HCl(aq) AlCl3(aq) + H2(g)2 6 2 3Now lets get organized. Write the information below the substances.3.45 g? gramsGram to Gram Conversions

*Aluminum is an active metal that when placed in hydrochloric acid produces hydrogen gas and aluminum chloride. How many grams of aluminum chloride can be produced when 3.45 grams of aluminum are reacted with an excess of hydrochloric acid? Al(s) + HCl(aq) AlCl3(aq) + H2(g)2 6 2 33.45 g? gramsLets work the problem.= g AlCl33.45 g AlWe must always convert to moles.Now use the molar ratio.Now use the molar mass to convert to grams.17.0Units matchgram to gram conversions

*MolarityMolarity is a term used to express concentration. The units of molarity are moles per liter (It is abbreviated as a capital M)When working problems, it is a good idea to change M into its units.

*A solution is prepared by dissolving 3.73 grams of AlCl3 in water to form 200.0 mL solution. A 10.0 mL portion of the solution is then used to prepare 100.0 mL of solution. Determine the molarity of the final solution.What type of problem(s) is this?Molarity followed by dilution.Solutions

*A solution is prepared by dissolving 3.73 grams of AlCl3 in water to form 200.0 mL solution. A 10.0 mL portion of the solution is then used to prepare 100.0 mL of solution. Determine the molarity of the final solution.1st:= mol L3.73 g200.0 x 10-3 L0.1402nd:M1V1 = M2V2 (0.140 M)(10.0 mL) = (? M)(100.0 mL)0.0140 M = M2molar mass of AlCl3dilution formulafinal concentrationSolutions

*50.0 mL of 6.0 M H2SO4 (battery acid) were spilled and solid NaHCO3 (baking soda) is to be used to neutralize the acid. How many grams of NaHCO3 must be used?H2SO4(aq) + 2NaHCO3 2H2O(l) + Na2SO4(aq) + 2CO2(g)Solution Stoichiometry

*50.0 mL6.0 M? g Look! A conversion factor!50.0 mL of 6.0 M H2SO4 (battery acid) were spilled and solid NaHCO3 (baking soda) is to be used to neutralize the acid. How many grams of NaHCO3 must be used?H2SO4(aq) + 2NaHCO3 2H2O(l) + Na2SO4(aq) + 2CO2(g)Solution Stoichiometry=Our Goal

*50.0 mL6.0 M? g 50.0 mL of 6.0 M H2SO4 (battery acid) were spilled and solid NaHCO3 (baking soda) is to be used to neutralize the acid. How many grams of NaHCO3 must be used?H2SO4(aq) + 2NaHCO3 2H2O(l) + Na2SO4(aq) + 2CO2(g)Solution Stoichiometry=Our Goal= g NaHCO3H2SO450.0 mL1 molH2SO4NaHCO32 molNaHCO384.0 gmolNaHCO350.4

*Solution Stoichiometry:Determine how many mL of 0.102 M NaOH solution are needed to neutralize 35.0 mL of 0.125 M H2SO4 solution.First write a balancedEquation. ____NaOH + ____H2SO4 ____H2O + ____Na2SO4 2 1 2 1

*Solution Stoichiometry:Determine how many mL of 0.102 M NaOH solution is needed to neutralize 35.0 mL of 0.125 M H2SO4 solution.Now, lets get organized. Place numerical Information and accompanying UNITS below each compound. ____NaOH + ____H2SO4 ____H2O + ____Na2SO4 2 1 2 10.102 M? mL35.0 mLSince 1 L = 1000 mL, we can use this to save on the number of conversionsOur Goal

*Determine how many mL of 0.102 M NaOH solution is needed to neutralize 35.0 mL of 0.125 M H2SO4 solution.Now lets get to work converting. ____NaOH + ____H2SO4 ____H2O + ____Na2SO4 2 1 2 10.102 M? mL35.0 mL= mL NaOH H2SO435.0 mL H2SO40.125 mol 1000 mL H2SO4 NaOH2 mol1 mol H2SO41000 mL NaOH0.102 mol NaOH85.8Units Match Solution Stoichiometry:shortcut

*What volume of 0.40 M HCl solution is needed to completely neutralize 47.1 mL of 0.75 M Ba(OH)2?1st write out a balanced chemicalequationSolution Stoichiometry

*What volume of 0.40 M HCl solution is needed to completely neutralize 47.1 mL of 0.75 M Ba(OH)2?2HCl(aq) + Ba(OH)2(aq) 2H2O(l) + BaCl20.40 M47.1 mL0.75 M? mL= mL HClBa(OH)247.1 mL1 mol Ba(OH)2HCl2 mol0.40 mol HCl HCl1000 mL176Units matchSolution Stoichiometry

*Solution Stochiometry Problem:A chemist performed a titration to standardize a barium hydroxide solution. If it took 23.28 mL of 0.135 M hydrochloric acid to neutralize 25.00 mL of the barium hydroxide solution, what was the concentration of the barium hydroxide solution in moles per liter (M)?First write a balanced chemical reaction.____HCl(aq) + ____Ba(OH)2(aq) ____H2O(l) + ____BaCl2(aq)2 1 2 1 23.28 mL0.135 mol L25.00 mL ? mol L

*Solution Stochiometry Problem:A chemist performed a titration to standardize a barium hydroxide solution. If it took 23.28 mL of 0.135 M hydrochloric acid to neutralize 25.00 mL of the barium hydroxide solution, what was the concentration of the barium hydroxide solution in moles per liter (M)?____HCl(aq) + ____Ba(OH)2(aq) ____H2O(l) + ____BaCl2(aq)2 1 2 1 23.28 mL0.135 mol L25.00 mL ? mol L = mol Ba(OH)2 L Ba(OH)225.00 x 10-3 L Ba(OH)2Units Already Match on Bottom!0.0629Units match on top!

*48.0 mL of Ca(OH)2 solution was titrated with 19.2 mL of 0.385 M HNO3. Determine the molarity of the Ca(OH)2 solution.

We must first write a balanced equation.

Solution Stochiometry Problem:

*48.0 mL of Ca(OH)2 solution was titrated with 19.2 mL of 0.385 M HNO3. Determine the molarity of the Ca(OH)2 solution. Ca(OH)2(aq) + HNO3(aq) H2O(l) + Ca(NO3)2(aq)2248.0 mL19.2 mL0.385 M= mol(Ca(OH)2) L (Ca(OH)2)19.2 mLHNO348.0 x 10-3L? Munits match!0.0770Solution Stochiometry Problem:

*Limiting/Excess/ Reactant and Theoretical Yield Problems :Potassium superoxide, KO2, is used in rebreathing gas masks to generate oxygen.4KO2(s) + 2H2O(l) 4KOH(s) + 3O2(g) a. How many moles of O2 can be produced from 0.15 mol KO2 and 0.10 mol H2O?b. Determine the limiting reactant.4KO2(s) + 2H2O(l) 4KOH(s) + 3O2(g) First copy down the the BALANCED equation!Now place numerical the information below the compounds.

*Limiting/Excess/ Reactant and Theoretical Yield Problems :Potassium superoxide, KO2, is used in rebreathing gas masks to generate oxygen.4KO2(s) + 2H2O(l) 4KOH(s) + 3O2(g) a. How many moles of O2 can be produced from 0.15 mol KO2 and 0.10 mol H2O?b. Determine the limiting reactant.4KO2(s) + 2H2O(l) 4KOH(s) + 3O2(g) 0.15 mol0.10 mol? molesTwo starting amounts? Where do we start?Hideone

*Limiting/Excess/ Reactant and Theoretical Yield Problems :Potassium superoxide, KO2, is used in rebreathing gas masks to generate oxygen.4KO2(s) + 2H2O(l) 4KOH(s) + 3O2(g) a. How many moles of O2 can be produced from 0.15 mol KO2 and 0.10 mol H2O?b. Determine the limiting reactant.4KO2(s) + 2H2O(l) 4KOH(s) + 3O2(g) 0.15 mol0.10 mol? molesHide Based on:KO2

= mol O20.15 mol KO20.1125

*Potassium superoxide, KO2, is used in rebreathing gas masks to generate oxygen.4KO2(s) + 2H2O(l) 4KOH(s) + 3O2(g) a. How many moles of O2 can be produced from 0.15 mol KO2 and 0.10 mol H2O?b. Determine the limiting reactant.4KO2(s) + 2H2O(l) 4KOH(s) + 3O2(g) 0.15 mol0.10 mol? molesBased on:KO2

= mol O20.15 mol KO20.1125HideBased on: H2O= mol O20.10 mol H2O0.150Limiting/Excess/ Reactant and Theoretical Yield Problems :

*Limiting/Excess/ Reactant and Theoretical Yield Problems :Potassium superoxide, KO2, is used in rebreathing gas masks to generate oxygen.4KO2(s) + 2H2O(l) 4KOH(s) + 3O2(g) a. How many moles of O2 can be produced from 0.15 mol KO2 and 0.10 mol H2O?Determine the limiting reactant.4KO2(s) + 2H2O(l) 4KOH(s) + 3O2(g) 0.15 mol0.10 mol? molesBased on:KO2

= mol O20.15 mol KO20.1125Based on: H2O= mol O20.10 mol H2O0.150What is the theoretical yield? Hint: Which is the smallest amount? The is based upon the limiting reactant?It was limited by theamount of KO2.H2O = excess (XS) reactant!

*Theoretical yield vs. Actual yieldSuppose the theoretical yield for an experiment was calculated to be 19.5 grams, and the experiment was performed, but only 12.3 grams of product were recovered. Determine the % yield.Theoretical yield = 19.5 g based on limiting reactantActual yield = 12.3 g experimentally recovered

*4KO2(s) + 2H2O(l) 4KOH(s) + 3O2(g) If a reaction vessel contains 120.0 g of KO2 and 47.0 g of H2O, how many grams of O2 can be produced?4KO2(s) + 2H2O(l) 4KOH(s) + 3O2(g) 120.0 g47.0 g? gHide oneBased on:KO2= g O2 120.0 g KO240.51Limiting/Excess Reactant Problem with % Yield

*4KO2(s) + 2H2O(l) 4KOH(s) + 3O2(g) If a reaction vessel contains 120.0 g of KO2 and 47.0 g of H2O, how many grams of O2 can be produced?4KO2(s) + 2H2O(l) 4KOH(s) + 3O2(g) 120.0 g47.0 g? gBased on:KO2= g O2 120.0 g KO240.51Based on:H2O= g O2Question if only 35.2 g of O2 were recovered, what was the percent yield?Hide47.0 g H2O125.3Limiting/Excess Reactant Problem with % Yield

*If a reaction vessel contains 120.0 g of KO2 and 47.0 g of H2O, how many grams of O2 can be produced?4KO2(s) + 2H2O(l) 4KOH(s) + 3O2(g) 120.0 g47.0 g? gBased on:KO2= g O2 120.0 g KO240.51Based on:H2O= g O247.0 g H2O125.3Determine how many grams of Water were left over.The Difference between the above amounts is directly RELATED to the XS H2O.125.3 - 40.51 = 84.79 g of O2 that could have been formed from the XS water.= g XS H2O84.79 g O231.83

*Calculate the molarity of a solution prepared by dissolving 25.6 grams of Al(NO3)3 in 455 mL of solution.After you have worked the problem, click here to see setup answer Try this problem (then check your answer):

*Chemistry: StoichiometryUnifying Theme: Constancy, Change, and MeasurementWhyWhatHow

Enduring UnderstandingsEssential Questions Major Indicators, PrioritizedAssessment ExamplesThe mole is basic unit of measurement in chemistry contains 6.02 x 1023 particles.The mole is the bridge between microscopic and macroscopic world and is used in unit analysis.Stoichiometry is used to predict quantities in chemical reactions.Products are limited by the amount of reactants such as natural resources and the limitation can be predicted.Mass of 1 mole substance can be determined based on the atomic mass in the Periodic Table.How is the mole used in chemical calculations?How is quantity of materials predicted in a chemical reaction?How do available materials limit the amount of products?How is molar mass calculatedEssentialHS4.4.5.1 use the coefficients of a balanced equation to predict amounts of reactants and products.HS4.4.5 demonstrate that adjusting quantities of reactants will affect the amounts of products formed.HS4.4.A.A identify and define Avogadros number and the mole concept operationally and conceptually.HS4.4.A.B utilize dimensional analysis to perform mole to mole, mass to mass, particle to mole, and mole to particle calculations.HS4.4.A.D use formula mass to calculate percent composition of a compound.HS.4.4A.F solve problems involving quantitative relationships in equations including stoichiometric concepts of mole to mole, particle to particle, mass to mass, mole to volume and volume to mole.HS4.4.A.G manipulate the limiting reagent concept qualitatively to conclude that the starting materials of the chemical industry such as petroleum, are limited resources and decisions must be made about their wise consumption.HS1.6.4 manipulate quantities and/or numerical values in algebraic equations.HS1.7.4 recognize mathematics as an integral part of the scientific process.HS4.4.1.5 calculate the molecular weight of a compound given in the Periodic Table. Pre-assessmentBrainstorm the collective terms such as dozen, pairs, gross, etc, and conversions between the terms and their numerical values. Formative AssessmentDevelop a creative concept map to illustrate the relationships among mole, mass, volume and number of particles. Summative AssessmentCompetition: Students are given a specific amount of iron filing and copper sulfate. They are required to return exact one gram of elemental copper after the reaction is completed.Write an article for your school paper about the air pollution in your community. Be sure to include the stoichiometric analysis.

*Chemistry: StoichiometryHow

Suggested Instructional SequenceDifferentiation ExamplesIntroduction of mole concept. - What is mole?- Determine the number of rice grains in a given container.Guided practice in calculation of atomic mass, molar mass, and formula mass.Utilize dimensional analysis to convert among mole, gram, volume and number of particles.- Lab Galvanized Iron: determine number of atoms of Zn coated on a piece of galvanized iron.- Guided practice: mole conversions using mole concept map.Percent of composition;- Lab percent of water in popcorn- Lab determining an Empirical formula such as copper sulfate hydrates.Introduction of Stoichiometry - Demonstration: electrolysis of water to illustrate the mole ratio.- Lab Use stoichiometric relationship to analyze the amount of silver replaced from silver nitric solution with the copper.Introduction of limiting reagent- Using manipulation to demonstrate limiting reagents (cake recipe)- A research project about the limitation of a natural resource, such as petroleum, copper.ChallengeDesign a lab to synthesize a quantitative product when given a variety of reactantsDesign a lab to determine the empirical formula for a given hydrate compound.Adjust Predict the volume of carbon dioxide that would be produced by adding specified amounts of vinegar to various teaspoons of baking soda. Carry out the activity and compare the experimental results with your prediction. Account for any differences in volume.ESOLCreate a PowerPoint presentation to explain stoichiometry