What Really Happens in Stoichiometry, Most of the Time Reactants are almost never mixed in the exact amounts necessary for them all to react. Therefore, some reacts are left over. Some reactant becomes used up and cannot go on to react with another reactant, causing that reactant to be left behind. The reactant that is used up first is THE LIMITING REAGENT. Example: The vending machine problem that takes only certain types of change.

Strategy for Identifying the Limiting Reagent Select a product, any product, of the balanced chemical reaction. Use the balanced chemical reaction equation to calculate the amount of the selected product that would be formed if the entire supply of first reagent were consumed. Repeat the process for the second reagent, the third reagent, etc. Identify the limiting reactant by which reactant gives the least amount of the selected product.

The Concept of Limiting Supplies My recipe for a bacon double cheeseburger is: 1 hamburger bun 2 hamburger patties 2 slices of cheese 4 strips of bacon B+2H+2C+4SBDC How many bacon double cheeseburgers can you make if you start with: 10 buns, 20 patties, 2 slices of cheese, 40 strips of bacon? 10 1

Example of Finding the Limiting Reagent Question: A reaction proceeds to completion by using 3 moles of A, 2 moles of B and 1 mole of C to form 5 moles of D. Given that you start with 7 moles of A, B, and C, what is the limiting reagent? Solution: We begin by writing a balanced chemical reaction: 3 A +2 B + C → 5 D Every mole of A produces 5/3 moles of D. Hence from 7 moles of A, we get 7 x 5/3 moles = 35/3 moles = 11.67 moles of D.

Example of Finding the Limiting Reagent Continued We repeat the same calculation for reagents B and C. For 7 moles of B we form 7 x 5/2 moles 35/2 moles =17.5 moles of D. For 7 moles of C we form 7 x 5 moles = 35 moles of D. Conclusion: Reagent A is the limiting reagent, which if you think more about it, is actually quite obvious from the start.

Calculating the Percentage Yield Percentage yield of product = [Actual yield of product]/[Theoretical yield of product] x 100 The theoretical yield equals the amount of product when all the limiting reagent reacts.

Following the Progress of a Reaction Make a table called the table of changes. Each column has three lines. The first is the initial chemical amount The second is the change in the chemical amount The third is the final chemical amount once this change has occurred.

Example: Table of Change Cu(s) +4 HNO3(aq)→Cu(NO3)2+2 NO2(g)+2 H2O(l) Cu HNO3 Cu(NO3)2 NO2 H2O Initial a b Change -x -4x x 2x Final a-x b-4x Discuss the units.

Reactions in Solution Solute Solvent Concentration Molarity = moles of solute per liter of solvent M = mol L-1 Strategy for solvent solution stoichiometry problems: Think again in terms of amounts (moles) and multiply concentration in M by volume in L.

How to Prepare a Solution The most common method of preparing a solution of known molar concentration is to weigh out an appropriate mass of solute, place it in a volumetric flask, and dilute to the mark with dissolution of the solute. The solute must dissolve before the final amount of the solvent is added, because the volume change on dissolution is not known. This procedure for preparing a molar solution can only be used when the pure solute is in an easily weighable form.

Appropriate instructions for the preparation of 500 mL of 0 Appropriate instructions for the preparation of 500 mL of 0.375 molar aqueous sodium acetate solution can be derived as follows: The molar mass of CH3COONa, sodium acetate, is 82.0245 g/mol. The mass of sodium acetate required would be (0.375 mol/L)(82.0245 g/mol)(0.500 L), which is 30.759 g. The procedure is then to place this mass in a 500 mL volumetric flask and add water to prepare the solution, with dissolution of the sodium acetate.

The simplest way to prepare a solution of known molarity is by dilution of a previously prepared solution of greater molar concentration. The preparation requires calculation of the amount of solvent to be added to a solution to prepare the desired solution.

Sample Calculation An aqueous solution which is 0.50 molar is available and 1.00 L of 0.10 molar aqueous solution of the solute is required. Such a solution can be prepared as follows: The amount of solute in 1.00 L of 0.10 molar aqueous solution is 0.100 mol. This amount of solute would be contained in 0.100 mol/(0.50 mol/L) = 0.2 L or 200 mL. The appropriate procedure would then be to take 200 mL of the original solution and dilute it to 1000 mL with the solvent, water.

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