 # Stoichiometry Chapter 5. Stoichiometry Quantitative relationships between reactants and products The balanced chemical equation gives us the relationships.

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Stoichiometry Chapter 5

Stoichiometry Quantitative relationships between reactants and products The balanced chemical equation gives us the relationships in moles Consider: N 2 + 3H 2  2NH 3 Three mol-mol conversion factors or mole ratios can be derived from balanced equation

Stoichiometry N 2 + 3H 2  2NH 3 One mol of N 2 produces 2 mol of NH 3 One mol of N 2 reacts with 3 mol of H 2

Stoichiometry 3 mol of H 2 produces 2 mol of NH 3 Inverse can also be used as a conversion factor

Example How many moles of NH 3 can be produced from 33.6 g of N 2 ? Convert grams of N 2 to mol of N 2, then convert mol of N 2 to mol of NH 3

Stoichiometry - Procedure 1.Write down what is given and what is requested in the problem 2.a. if a mass is given, use the molar mass to convert mass to moles of what is given b). if a number of molecules is given, use Avogadro’s number to convert to moles of what is given

Stoichiometry procedure 3.Using the correct mole ratio from the balanced equation, convert moles of what is given to moles of what is requested 4.a. If a mass is required, convert moles of what is requested to mass of what is requested b. If a number of molecules is requested, use Avogadro’s number to convert to numbers of molecules of what is requested 5.Procedure is summarized on next slide

Stoichiometry

Limiting reactant If specific amounts of each reactant are mixed, the reactant that produces the least amount of product is called the limiting reactant Think of hot dogs and buns –hot dogs are sold in packs of ten –hot dog buns are sold in packs of eight –how many hot dog-hog dog bun combinations can you make with one pack of hot dogs and one pack of hot dog buns?

Limiting reactant considerations When reactants are mixed in exactly the mass ratio determined from the balanced equation, the mixture is said to be a stoichiometric mixture Example: 4.0 g H 2 + 32.0 g O 2  36.0 g H 2 O Other mass ratios require calculations to determine the limiting reactant

Limiting reactant procedure Convert amount of each reactant to the number of moles of product using mole ratios The limiting reactant is the one that produces the smallest amount of product

Limiting reactant example 2 CH 3 OH (l) + 3 O 2 (g)  2 CO 2 (g) + 4 H 2 O (g) Mix 40.0 g of methanol with 40.0 g of O 2, what is the mass of CO 2 produced? Methanol, CH 3 OH, has a molar mass of 40.0 g O 2 has a molar mass of 32.00 g

Process

Percent yield The actual yield is the amount of products obtained when the reaction is run The theoretical yield is the calculated amount of product that would be obtained if all of the limiting reactant was converted to a given product The percent yield is the actual yield in grams or moles divided by the theoretical yield in grams or moles times 100%

Incomplete conversion In some cases, a reverse reaction occurs whereby reactants are reformed from products This limits the percent of reactants that are converted to products Such reactions are known as reversible reactions The reaction between N 2 and H 2 to produce NH 3 is a reaction which is reversible. This has severe consequences for the commercial production of ammonia

Example calculation For the conversion of N 2 and H 2 to NH 3 –4.70 g H 2 react with N 2 –12.5 g of NH 3 is formed The theoretical yield is 26.5 g of NH 3 in this reaction 47.5% (commercially the yield is only 28%)

Working with Solutions Solution Concentration: Molarity The amount of solute dissolved in a given solvent reported as moles of solute per liter of solution. Calculation of molarity is done by calculating the moles of solute present and dividing by the total volume of the solution in liters.

Preparing Solutions of Known Concentration Begin with pure solute Example: Make a 1.5 M solution of NaOH Molecular weight of NaOH: Na = 24; O = 16; H = 1 Total = 40 1.5 *40 = 60 g NaOH / liter final volume. Dissolve 60 g NaOH in 400 mL distilled water and dilute to l liter final volume. Mix well.

Begin with concentrated solution (dilution calculation - MV = MV) Example: Make 100 mL 0.05 M NaOH from a 1.5 M solution. 0.05 * 100 = 1.5 * ? (0.05 * 100 )/1.5 = 3.33 mL of 1.5 M NaOH diluted to 100 mL with distilled water

Stoichiometry of Reactions in Aqueous Solution 1. Write the balanced equation 2. Calculate moles from masses or moles from molarity 3. Use a stoichiometric factor 4. Calculate mass from moles or molarity from moles and volume.

Titrations Essence of any technique of quantitative chemical analysis - determination of the quantity of a given constituent in a mixture -If you know the balanced equation for the reaction and the exact quantity of one of the reactants, then you can calculate the quantity of any other substance consumed or produced in the reaction.

Titration Titration - procedure that allows for the determination of an unknown by adding carefully measured quantities of one solution into another solution when the exact concentration or amount of one of the solutions is known.

Indicator - dye that changes color when the solution used for analysis if complete Buret - a measuring cylinder that most commonly has a volume of 50.0 mL and is calibrated in 0.1 mL divisions Equivalence point - the point at which the reaction is complete.

Pre Lab Analysis of a Commercial Bleach Use titration to determine the molarity of the NaOCl as well as the percent by mass of the NaOCl in a sample of bleach using sodium thiosulfate as the titrant and iodine as the indicator.

Reactions 1. Acidified iodide ion is added to hypochlorite ion solution, and the iodide is oxidized to iodine. 2 H+(aq) + ClO - (aq) + 2 I - (aq)  Cl - (aq) + I 2 (aq) + H 2 O ( l) 2. Iodine is only slightly soluble in water. In an aqueous solution of iodide ion, iodine dissolves very readily. The triiodide ion forms in this situation. Triiodide is a combination of a neutral I2 molecule with an I- ion. The triiodide ion is yellow in dilute solution, and dark red-brown when concentrated. I 2 (aq) + I-(aq)  I 3 -(aq) 3. The triiodide is then titrated with a standard solution of thiosulfate ions, which reduces the iodine back to iodide ions: I 3 -(aq) + 2 S 2 O 3 2- (aq)  3 I - (aq) + S 4 O 6 2- (aq)

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