Stoichiometry

292 Views

Download Presentation
## Stoichiometry

- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -

**Introduction to**Stoichiometry Mr. Shumway Chemistry 1**Stoichiometry**• "Stoichiometry" is derived from the Greek words • στοιχεῖον (stoicheion, meaning element]) and • μέτρον (metron, meaning measure. • That’s why it’s a funny looking word**Stoichiometry**• Nevertheless, • Stoichiometry is the science that deals with measuring the various elements in chemical reactions**Stoichiometry**• Write down all that you can about this reaction below N2(g) + 3H2(g) 2 NH3(g)**Stoichiometry – Interpreting Chemical Equations**• What do you look at when looking at a chemical equations? There is a lot of information in a chemical equation… • Particles • Moles • Mass • Volume**Stoichiometry**N2(g) + 3H2(g) 2 NH3(g) Particles • How many particles are there for each reactant? • How many particles are there in the product? • There is 1 molecule of Nitrogen reacting with 3 molecules of hydrogen gas • Producing 2 molecules of ammonia • What is the mole ratio for this reaction? • The mole ratio for this reaction is 1:3:2**Stoichiometry**N2(g) + 3H2(g) 2 NH3(g) H H N H H N N H H H H H H H N H**Stoichiometry**N2(g) + 3H2(g) 2 NH3(g) Moles • How many moles are there for each reactant? • How many moles are there in the product? • There is 1 mol of Nitrogen reacting with 3 mol of hydrogen gas • Producing 2 mol of ammonia • What is the mole ratio for this reaction? • The mole ratio for this reaction is 1:3:2**Stoichiometry**N2(g) + 3H2(g) 2 NH3(g) Mass – Law of conservation of mass • What is the mass of each reactant? • What is the mass of each product? • Ex • 1 mol of N2 = 28.0 grams of N2 • 1 mol of H2 = 2.0 grams of H2 , therefore, 3 mol of H2 = 6.0 grams of H2 • 1 mol of NH3= 17 grams, therefore, 2 mol of NH3 = 34.0 grams • The mass of the reactants = the mass of the products 28.0 + 6.0 = 34.0**Stoichiometry**N2(g) + 3H2(g) 2 NH3(g) Volume • 1 mol of a gas at STP = 22.4L • Therefore, what are the volumes of each gas from the reactants? • What is the volume of the product gas that is formed? • Ex • 22.4 L of N2 reacts with 67.2L of H2 (3mol x 22.4L) • Forming 44.8L of ammonia**Stoichiometry**Let’s Look at this problem 2H2S(g) + 3O2(g) 2SO2(g) + 2H2O(g)**Stoichiometry – Individual Work**• Balance the following equation and write what down all the information that you can • Number of particles • Number of moles of reactants and products • Mass of reactants and products • Volumes of gases at STP (if applicable) • C2H2(g) + O2(g) CO2(g) + H2O(g) • Na(s) + H2O(l)NaOH(aq) + H2(g) • CO(g) +O2(g) CO2(g)**DRILL**• What do the coefficients tell you in a chemical reaction? • What can the coefficients tell you?**DRILL**• What do the coefficients tell you in a chemical reaction? • The coefficients tell you the number of molecules/atoms there are in a chemical reaction. • What can the coefficients tell you? • They can tell you the mole ratios between each of the reactants and products**Mole to Mole Ratios**Stoichiometry Mr. Shumway Chemistry 1**Stoichiometry**We can use the information from the chemical equation and use them to help us calculate useful information • Mole – Mole Calculations • Mass – Mass Calculations • Mass-mole and mole mass conversions**Stoichiometry**• Mole – Mole Calculations • Knowing the amount of moles of reactants, we can calculate the number of moles of product • Using the mole ratios of the chemical equation**Stoichiometry**• Mole – Mole Calculations How many moles of ammonia are produced when 0.60 mol of nitrogen reacts with hydrogen? N2(g) + 3H2(g) 2NH3(g) This is what we want This is what we have**Stoichiometry**How many moles of ammonia are produced when 0.60 mol of nitrogen reacts with hydrogen? N2(g) + 3H2(g) 2NH3(g) 1 What is the ratio/relationship between these molecules?**Stoichiometry**How many moles of ammonia are produced when 0.60 mol of nitrogen reacts with hydrogen? N2(g) + 3H2(g) 2NH3(g) 2 mol of NH3(g) = 1.2 mol NH3 0.60 mol of Nitrogen X 1 mol of N2(g) Write down what you have first Because we have the number of moles, we need a conversion factor that has mol of Nitrogen in it “A mole ratio”**Stoichiometry**Let’s look at this problem on the board • Calculate the number of moles of reactants required to make 7.24 mol of Ammonia N2(g) + 3H2(g) 2NH3(g)**Stoichiometry - Individual Work**The formation of aluminum oxide from its constituent elements is represented by this equation • How many moles of aluminum are needed to form 2.3 mol of Al2O3? • How many moles of oxygen are required to react completely with 0.84 mol of Al? • Calculate the number of moles of Al2O3 formed when 17.2 mol of O2 reacts with aluminum.**Drill**• Design a step-by-step method that will help you make conversions between moles to moles in a chemical reaction.**Mole to Mass Calculations**Stoichiometry Mr. Shumway Chemistry 1**Stoichiometry**• Review • You can calculate the number of moles of product through the mole ratios of the chemical equation. • Following through with step by step dimensional analysis • If we can calculate the number of moles, we can also calculate the mass of the product and/or the reactants.**Stoichiometry**A + X AX Steps to Follow • It is important that if you are given grams to first convert to moles because you can use the mole ratio from the chemical equation • From moles of X to moles of AX (using the mole ratio) • Convert moles of AX to grams of AX**Stoichiometry**• Calculate the number of grams of NH3 produced by the reaction of 5.40g of hydrogen with nitrogen N2(g) + 3H2(g) 2NH3(g)**Stoichiometry**• Step 1: calculate the number of moles by the mass given N2(g) + 3H2(g) 2NH3(g) 1 mol of H2(g) = 2.7 mol H2 5.40 g of H2 gas X 2.0 g of H2(g) Write down what you have first Because we have the grams of Hydrogen, we need a conversion factor that has grams/mol with the proper molecules. Using the molar mass of the molecule.**Stoichiometry**• Step 2: Convert from moles to moles N2(g) + 3H2(g) 2NH3(g) 2 mol of NH3(g) = 1.8 mol NH3(g) 2.70 mol of H2 gas X 3 mol of H2(g) Write down what you got from the last problem Because we have the moles of Hydrogen, we need a conversion factor that will give us the number of moles of ammonia gas, using the mole ratio in the chemical reaction. Essentially, we’re converting from moles given to moles desired**Stoichiometry**• Step 3: Convert from moles to grams N2(g) + 3H2(g) 2NH3(g) 17.0 g of NH3(g) = 30.6 g NH3(g) 1.80 mol of NH3(g) X 1 mol of NH3(g) Write down what you got from the last problem Because we have the moles of ammonia, we need a conversion factor that will give us the grams of ammonia gas, using the molar mass of the molecule.**Stoichiometry – Teacher**The combustion of acetylene gas is represented by this equation. 2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(g) • How many grams of oxygen are required to burn 13.0 g of C2H2? • How many grams of CO2 and grams of H2O are produced when 13.0g of C2H2 reacts with the oxygen required to burn 13.0g of C2H2?**Stoichiometry – Individual Work**Acetylene gas, C2H2, is produced by adding water to calcium carbide, CaC2 CaC2(s) + 2H2O(l) C2H2 + Ca(OH)2(aq) • How many grams of acetylene are produced by adding water to 5.00g of CaC2? • How many moles of CaC2 are needed to react completely with 98.0 g of H2O?**DRILL**• Construct a flow chart/map of how to solve the problems you’ve encountered thus far.**Review**• Get into your table groups. • Each group will have a different problem to work on, you will have 20 min to solve the problem. Everyone in the group MUST KNOW HOW TO SOLVE THE PROBLEM • One random person will be chosen to come to the board and do the problem and explain their work, this will determine the groups grade.**Drill**• What is a limiting reagent that you described in your mini lab?**Limiting Reagent**Stoichiometry Mr. Shumway Chemistry 1**Stoichiometry – Limiting Reagent**• Limiting reagent • Limits or determines the amount of product that can be formed in a reaction • Excess reagent • Is more than enough to react with a limiting reagent**Stoichiometry – Limiting Reagent**Ex. Problem Sodium Chloride is prepared by the reaction of sodium metal with chlorine gas. What will occur when 6.70 mol of Na reacts with 3.20 mol of Cl2? What is the limiting reagent? How many moles of NaCl are produced? How much excess reagent remains unreacted?**Stoichiometry – Limiting Reagent**• Step 1- What do we know? • We know/have • 6.70 mol of Na • 3.20 mol of Cl2 • Step 2 – What do we want to know? • We want to calculate three things • What is the limiting reagent? • How many moles of NaCl are produced • How much excess reagent remains unreactedd**Stoichiometry – Limiting Reagent**• Step 3 – Write down a BALANCED chemical reaction • Step 4- convert what you know into moles if not already**Stoichiometry – Limiting Reagent**• Step 5 – Look at the mole ratio and calculate the amount of substance needed for a complete reaction • Use this information to determine the limiting reagent and excess reagent and label them.**Stoichiometry – Limiting Reagent**Start with what we know Required amount of Chlorine Mole Ratio • 1 mol of Cl2 • = 3.35 mol of Cl2 • 6.70 mol of Na x • 2 mol of Na Knowing that we have 6.70 mol of Na and knowing the mole ratio between sodium and chlorine, we can calculate the minimum amount of chlorine we need for this reaction to take place Using the mole ratio, we can see that 3.35 mol of Cl2 is needed to react with 6.70 mol of Na Because we do not have enough Cl2, Cl2 is the limiting reagent. How much Chlorine did the problem say we had? Can someone explain why? 3.20 mol of Cl2!**Stoichiometry – Limiting Reagent**Mole Ratio Because Cl2 is the limiting reagent, the amount of product is determined by the amount of Cl2 we have. Total amount of product formed Therefore, the number of moles of Cl2 is going to be used to calculate the maximum amount of product that can be formed. HOW MUCH EXCESS DO WE HAVE? ( Na?)**Stoichiometry – Limiting Reagent**Mole Ratio The amount of excess reagent remaining is the difference between the given amount (in the beginning of the problem, a value of 6.70 mol of Na) and the amount of sodium needed to react with the limiting reagent. Amount of Na used up in the reaction 6.70 mol Na – 6.40 mol Na = 0.30 mol Na in excess Therefore, we need to now calculate how much Na would be used up in the reaction with Cl2**Stoichiometry: L.R. Teacher Demo**Copper reacts with sulfur to form copper (I) sulfide. What is the maximum number of grams of Cu2S that can be formed when 1.87 mol of Cu reacts with 2.19 mol of S?**Stoichiometry: l.r. Individual Work**Hydrogen gas can be produced in the laboratory by the reaction of magnesium metal with hydrochloric acid. How many grams of hydrogen can be produced when 1.45 mol of HCl is added to 2.31mol of Mg? Assuming STP, what is the volume of this hydrogen?**Stoichiometry – Limiting Reagent**Write down a systematic approach to dealing with Limiting reagent type problems. How would you solve this type of problem, step-by step?**Drill**• What is a limiting reagent? How can you know which reactant is the limiting reagent? • What is an excess reagents? How can you know which reactant is the excess reagent?