Deals with amounts of reactants used & products formed. 11.1 STOICHIOMETRY Deals with amounts of reactants used & products formed.

What is Stoichiometry? The study of the relationship between amounts of reactants used and products formed in a chemical reaction Based on the Law of Conservation of Mass and Matter Matter is neither created nor destroyed Mass of reactants equals the mass of the products

MOLE-MASS RELATIONSHIPS Balanced equation: 4Fe (s) + 3O2(g)  2Fe2O3(s) Interpret: 4 atoms 3 molecules 2 formula units (Particles) Mole ratio: 4 moles 3 moles 2 moles (Coefficients) Note: Particles of ionic compounds are called “formula units” Calculate the mass of each reactant and product by multiplying the number of moles by the molar mass

SHOW MASS IS CONSERVED 4Fe (s) + 3O2(g)  2Fe2O3(s) Mass reactants: 4 mol Fe (55.8g Fe) = 223.2 g Fe (1 mole Fe) 3 mol O2 (32.00 g O2) = + 96.0 g O2 (1 mol O2) ____________ 319.2 g Total

MASS OF PRODUCTS: 4Fe (s) + 3O2(g)  2Fe2O3(s) 2 mol Fe2O3 (159.6 g Fe2O3) = 319.2 g (1 mol Fe2O3) Equals the mass of the reactants (319.2g) Law of Conservation of Matter

PROBLEM Interpret in terms of particles, moles, and mass. Show that mass is conserved: (Hint: look at coefficients for particles & moles) 4 Al + 3O2  2 Al2O3 Particles: Moles: Mass: Conserved?

SOLUTION: 4Al + 3O2  2 Al2O3 Law of Conservation of Matter shown. Particles: 4 molecules 3 molecule 2 molecule Moles: 4 mole 3 mole 2 moles Mass: 4 (27.0 g) + 3 (32.0 g) = 2 (102.0 g) Conserved? 204.0 g = 204.0 g YES! Law of Conservation of Matter shown.

There are six mole ratios for the following: MOLE RATIO – is a ratio between the number of moles of any two substances in a balanced chemical equation There are six mole ratios for the following: Ex. 4 Al(s) + 3 O2(g)  2 Al2O3(s) Note: 4 moles 3 moles 2 moles 4 mol Al and 3mol O2 3 mol O2 4 mol Al

4 mol Al and 2 mol Al2O3 2 mol Al2O3 4 mol Al 3 mol O2 and 2 mol Al2O3 MOLE RATIO – is a ratio between the number of moles of any two substances in a balanced chemical equation 4 mol Al and 2 mol Al2O3 2 mol Al2O3 4 mol Al 3 mol O2 and 2 mol Al2O3 2 mol Al2O3 3 mol O2 All stoichiometry calculations begin with a balanced equation and mole ratios!!

PROBLEM - FIND MOLE RATIOS FOR: 2 NH3  N2 + 3 H2 2 mol 1 mol 3 mol Ans. 2 mol NH3 or 1 mol N2 1 mol N2 2 mol NH3 2 mol NH3 or 3 mol H2 3 mol H2 2 mol NH3

Ans. 2 NH3  N2 + 3 H2 1 mole N2 or 3 mole H2 3 mole H2 1 mole N2

12.2 STOICHIOMETRIC CALCULATIONS There are 3 Basic Stoichiometry Calculations

1. Mole to Mole Conversions A piece of magnesium burns in the presence of oxygen forming magnesium oxide (MgO). How many moles of oxygen are needed to produce 12 moles of magnesium oxide? Step 1: Write a balanced equation 2 Mg (s) + O2 (g)  2 MgO (s)

Write mole ratios Choose the correct mole ratio needed for this problem

Mole to Mole Conversion cont’d Multiply the known number of moles of MgO by the mole ratio 6 mols of oxygen is needed to produce 12 mols of magnesium oxide

2. Mole to Mass Conversions The following reaction occurs in plants undergoing photosynthesis CO2(g) + H2O(l) C6H12O6(s)+ O2(g) How many grams of glucose (C6H12O6) are produced when 24.0 mols CO2 reacts in excess water?

Mole to Mass Conversion cont’d Write a balanced equation 6 CO2(g) + 6 H2O(l)  C6H12O6(s) + 6 O2(g) Use mole ratios to determine the number of moles of glucose produced by the given amount of carbon dioxide

Mole to Mass Conversion cont’d Multiply by the molar mass 721 g glucose is produced from 24.0 moles carbon dioxide

3. Mass to Mass Problems The only new step is first step: Convert grams of given substance to moles! Massgiven  Molesgiv Molesdesired Massdes ÷ molar massgiven X mole ratio X molar massdesired 3 step problem!

Label: 25.0 g ? g NH4NO3  N2O + 2H2O Mole Ratio: 1 mol 1 mol 2 mol Ammonium nitrate (NH4NO3) produces N2O gas and H2O when it decomposes. Determine the mass of water produced from the decomposition of 25.0 g of NH4NO3. 1) Find moles NH4NO3 (molar mass): 80.04 g/mol Use the inverse of the molar mass to convert grams of NH4NO3 to moles of NH4NO3

Label: 25.0 g ? g NH4NO3  N2O + 2H2O Mole ratio: 1 mol 1 mol 2 mol Determine the mole ratio of mol H2O to mol NH4NO3 from the chemical equation. The desired substance is the numerator. Multiply mol NH4NO3 by the mole ratio. Calculate the mass of H2O using the molar mass.

Label: 5.6g ? g PROBLEM: N2 + 3H2  2 NH3 If 5.6 g nitrogen reacts completely with hydrogen, what mass of ammonia is formed? 1) Find moles of nitrogen (given) – molar mass: 5.6 g N2_____ mass N2 = 2 x 14.01 = 28.02 g/mol 1 g N2 fill in units 5.6 g N2_1mol N2 1 28.02 g N2 fill in the rest

Label: 5.6 g ? g N2 + 3H2  2 NH3 Mole Ratio: 1mol 3mol 2mol 2) Find moles ammonia (desired) - mole ratio: 5.6 g N2 1mol N2 _2 mol NH3___ 1 28 g N2 1 mol N2 3) Find grams ammonia (desired) – molar mass: NH3 = 14 + 3 = 17 g/mole 5.6 g N2 1mol N2 _2 mol NH3 17.0 g NH3 1 28g N2 1 mol N2 1 mol NH3 given ÷ MM given mole ratio x MM desired = 6.8 g NH3 Mass H2 = 6.8 g – 5.6 g = 1.2 g LOCOM

12.3 LIMITING REACTANTS

What is a Limiting Reactant? A limiting reactant is the reactant that limits (stops) a reaction and determines the amount of product An excess reactant is any reactant that is left over after the reaction stops (all reactants except the limiting reactant)

Ingredients: You don’t always have the exact amounts. IN ORDER FOR CHEMICAL REACTION TO OCCUR, YOU MUST HAVE A COMPLETE SET OF REACTANTS (REAGENTS): You don’t always have the exact amounts. Ex. Let’s make some McBurgers!!! Ingredients: 2 buns; 1 burger patty; 1cheese slice; 1 tomato slice; 1 lettuce leaf; 3 pickles

How many McBurgers can you make? RECIPE CALLS FOR: 2 buns 1 burger patty 1 cheese slice 1 tomato slice 1 lettuce leaf 3 pickles YOU HAVE AVAILABLE: 6 buns 3 burger patties 5 cheese slices 6 tomato slices 5 lettuce leafs 6 pickles How many McBurgers can you make?

Based on the individual ingredients you have available: You have enough: Buns for (3) Burger Patties for (3) Cheese Slices for (5) Tomato Slices for (6) Lettuce Leafs for (5) Pickles for (2)

(Remember each McBurger requires 3 pickles) The Limiting Reactant (LR) for the McBurger making process in the pickles This is the ingredient we ran out of first and were unable to continue with the McBurger making process We were only able to make 2 McBurger from the 6 pickles we had available to use (Remember each McBurger requires 3 pickles)

Excess reactants (XR) are the ingredients not used in the McBurger making process: 4 buns 1 burger patty 3 cheese slices 4 tomato slices 3 lettuce leafs Note: You made 2 McBurgers

PRACTICE PROBLEM: N2 + 3H2  2 NH3 Using the above equation you are given 3.0 mols N2 and 5.0 mols H2 Determine the limiting reactant Determine the excess reactant Determine the amount of NH3 produced

l Label: 3.0mol 5.0 mol ?mol N2 + 3H2  2 NH3 M.R. 1 mol 3 mol 2 mol Draw 2 bridges. Smaller product is your answer. (Always use L.R. to find answer!) 3.0 mol N2|_2 mol NH3 = 6.0 mol NH3 | 1 mol N2 (N2 is XS reactant) 5.0 mol H2|_2 mol NH3 = 3.3 mol NH3 | 3 mol H2 Ans./H2 is L.R.

l Label: 3.0mol 5.0 mol ?mol N2 + 3H2  2 NH3 M.R. 1 mol 3 mol 2 mol Alternative way: Compare 2 mole ratios! Note: 1mole N2 reacts with 3 mol H2 3.0 mol N2 reacts with 9.0 mol H2 Given – 5.0 mol H2 (Don’t have enough H2!) Thus H2 is LR. Use H2 to solve for NH3. 5.0 mol H2|_2 mol NH3 = 3.3 mol NH3 | 3 mol H2

ANSWER: 3. 3 mol ammonia formed. Hydrogen (H2) is L. R ANSWER: 3.3 mol ammonia formed. Hydrogen (H2) is L.R. Nitrogen (N2) is the XS.

IF A PAPER BURNS IN A ROOM: + What is the limiting reactant? What is in excess? What would happen if the paper burned in a closed jar? LR? XS? +

4.0mol 7.0mol PROBLEM: 2 Al + 3 Cl2  2 AlCl3 Given 4.0 mol Al and 7.0 mol Cl2, what is the maximum amount of aluminum chloride formed? Step 1: 2 bridges: 4.0 mol Al|______________ | mol Al 7.0 mol Cl2 |________ | mol Cl2

Label: 4.0mol 7.0 mol ? Mol 2 Al + 3 Cl2  2 AlCl3 4.0 mol Al|__2 mol AlCl3 = 4.0 mol AlCl3 | 2 mol Al 7.0 mol Cl2 |_2 mol AlCl3 = 4.7 mol AlCl3 | 3 mol Cl2 *Al is L.R. Ans. 4.0 mol AlCl3 *Cl2 is XS

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