Limiting Reagent Reagent = fancy word for “Reactant” Complete packet in it’s entirety. Complete procedure for the lab (last 2 pages of this packet). you will be doing this when you return from break. Make sure to check your homework solutions from the past two classes’ book problems Have a wonderful break!

Making a Sandwich… As a top employee at the local sandwich shop (potbelly? Subway?) you are very familiar with the balanced equation that explains how to make a simple sandwich… Balanced Equation for a simple sandwich 2 slc of Bread + 3 slc of Meat + 1 slc of Cheese  1 Sandwich

Making a Sandwich… 2 slc of Bread + 3 slc of Meat + 1 slc of Cheese  1 Sandwich You show up to work one day and find that you have the following amounts of the necessary ingredients: 46 slices of bread 52 slices of meat 19 slices of cheese Your boss wants to know how many sandwiches you can make with these ingredients…

Limiting Reagent To determine the amount of product that can be formed, we must figure out which reactant limits us (which reactant runs out first) What ever amount of product can be formed from the limiting reactant is the MAXIMUM amount of product that can be formed Work with your table to determine the maximum amount of sandwiches that can be made (then turn page over for answers)

Sandwich 2 slc of Bread + 3 slc of Meat + 1 slc of Cheese  1 Sandwich 46 slices of bread 52 slices of meat 19 slices of cheese 1 Sandwich 1 slc Cheese 3 slc Meat 2 slc Bread 46 slc bread x = 23 Sandwiches 52 slc meat x = 17 Sandwiches 19 slc cheese x = 19 Sandwiches

Limiting Reagent Our limiting reagent is the meat We can make a maximum of 17 sandwiches What will we have left over? If you make 17 sandwiches then you only need 34 pieces of bread. You started with 46. Therefore, you have 12 pieces of bread left over

Math for determining Leftovers If you make 17 sandwiches then you only need 34 pieces of bread. You started with 46 so you have 12 pieces of bread left over 17 Sandwiches x = 34 slc Bread Initial Bread – Bread Used = Leftover Bread 46 slc bread – 34 slc bread = 12 slc bread left 1 Sandwich 2 slc Bread

Procedure for Finding Limiting Reactant (LR) 1.Write and balance the equation for the rxn 2.Convert known masses of reactants into MOLES 3.Determine how many MOLES of product each reactant can make. 4.The smaller # of moles of product is the LR. 5.The smaller # of moles of product is the maximum amount of product that will form. 6.* If necessary, convert moles of product into GRAMS of product.

Practice #1 Try this: synthesis of ammonia. – Write the unbalanced equation for nitrogen gas reacting with hydrogen gas to form ammonia. – The coefficients are a ratio! So, balance the reaction:

Practice #1 (cont’d) If we have 6 ‘parts’ of hydrogen gas, how many ‘parts’ of ammonia would we make? – Another way to say that question is… – 6 mol H 2 = ? mol NH 3 Scenario: you have 2.8 mol of Nitrogen gas and 3.5 mol of Hydrogen gas. Which of the two reactants is the limiting reagent? (circle your limiting reagent)

Practice #1 (cont’d) 2.8 mol nitrogen gas = ? mol ammonia?

Practice #1 (cont’d) 3.5 mol hydrogen gas = ? mol nitrogen gas?

Practice #1 (cont’d) You just completed mol to mol calculations. – mol A  mol B But in lab, remember, we must use mass, so our T- tables include 2 more conversions. – g A  mol A  mol B  g B Problem: Say you check on your tanks of hydrogen and nitrogen and you have a full tank of hydrogen (excess hydrogen), and only 18.2 g of nitrogen gas. How much ammonia could you produce, in grams?

Practice #1 (cont’d) Complete this problem in your notes. Write the correct setup (triple step conversion) with units first! What is in excess (the product with greater yield)? ____________. Therefore, this is not the limiting reagent. The word “only” means what? – That Nitrogen gas is the: ______________ _____________

Limiting Reagent (Reactant) Rules Limiting reagent: The REACTANT that will limit the amount of product that can be made Maximum theoretical yield: The maximum amount of product (typically in grams) that can be produced due to the limiting reagent

Limiting Reagent (Reactant) Rules To solve these you will need to write out TWO t- tables – 1. Calculate the amount of product that can be made from the amount of reactant A – 2. Calculate the amount of the SAME product that can be made from the amount of reactant B Compare the two masses of product. The lower amount of product is the maximum amount you can form.

Problem #2 Many students wondered what would happen if an excess of 5.00g of acetic acid was added in to the reaction. Let’s see if acetic acid or sodium hydroxide is the limiting reagent. For example, you have 6.00 g of acetic acid and 3.5 g of sodium hydroxide. Which reactant is the limiting reagent? – Remember, water was evaporated, so we want to see which reactant will limit (or the one with the least) the product of sodium acetate

Problem #2 Balance the equation Convert 6.00 g of acetic acid to grams of sodium acetate (3 step conversion). – g A  mol A  mol B  g B Convert 3.5 g of sodium hydroxide to grams of sodium acetate (3 step conversion). – g A  mol A  mol B  g B Which reactant with their respective amounts in grams produces less product? That is the limiting reagent. (work in space below)

Problem #3 What mass of hydrogen is produced by the reaction of 4.73g of magnesium metal with 1.83g of water? (work in space below) (h2o) vs g (mg)

Problem #4 What mass of AgCl is produced when 4.22g of AgNO 3 react with 7.73g of AlCl 3 in a double displacement reaction? (work in space below) 3.56g (ag) vs g (al)

Limiting Reagent Pre-Lab (last 15min) Write the balanced equations for the reaction between Lead (II) nitrate and Potassium Iodide What evidence do we have that a reaction will occur? – Is Iodide soluble or insoluble with Pb 2+ ? – Is Nitrate soluble or insoluble with K + ?

Limiting Reagent Pre-Lab (last 15min) In the space below, write the following: – Balanced equation – Complete ionic reaction – Net ionic equation Below those three equations: – Goal: Create a procedure to completely isolate the solid from all other substances and determine the percent yield for your experiment. You may begin with anywhere between 0.50g and 1.00 of each reactant (can vary) – Do this now, otherwise you’ll forget over break!