The Mathematics of Chemical Reactions Chemical Stoichiometry Quantitative Relationships between the reactants and products in a chemical reaction

Understanding chemical reactions helps us to predict what will be produced by a reaction. We also need to know how much will be produced by a reaction. We also can determine how much reactant is required to produce a desired amount of product. Chemical Stoichiometry

Stoichiometry – Solving the problems: A simple Process - Step #1: Step #2: Step #4: Step #3: Balance the Equation. put the measurements or quantities that you were given over the reactant or product to which that value pertains. Solve the Problem. Equation on the Bottom. Problem on the Top. put the measurements or quantities from the equation under the reactant or product to which that value pertains. Cross multiply and divide. V ertical A greement

Mass – Mole Stoichiometry KClO 3 ==> KCl + O 2 K = 1 x 39.1 = 39.1 Cl = 1 x 35.5 = 35.5 O = 3 x 16.0 = mol Given g of potassium chlorate, determine the moles of oxygen gas produced gX mol 122.6g (122.6) g g 300 g mol g = 1.22 mol

Determine the mass of iron which could be produced by the reaction of grams of iron(III) oxide with enough carbon to use it all. Fe 2 O 3 + C ==> Fe + CO gX g g g Fe = 2 x 55.8 = O = 3 x 16.0 = 48.0 (55.8) (159.6) g Mass – Mass Stoichiometry

Determine the volume of carbon dioxide produced by the reaction of L of ethane with enough oxygen to use it all. C 2 H 6 + O 2 ==> H 2 O + CO LX L LL Remember : Volume of a STP 1 mole = 22.4 L (22.4) 89.6 (22.4) L Volume in Stoichiometry 426

% Yield Problems What is the percent yield of salt in the following reaction when 50.0 g of sodium bicarbonate reacts with HCl, and your lab partner measured the salt produced in the evaporating dish at 28.6 g. NaHCO 3 + HCl ==> NaCl + H 2 O + CO 2 Na = 1 x 23.0 = 23.0 H = 1 x 1.0 = 1.0 C = 1 x 12.0 = 12.0 O = 3 x 16.0 = g 50.0 g 58.5 g x g Na = 1 x 23.0 = 23.0 Cl = 1 x 35.5 = X 100 % = 82.2 % 34.8 g

Limiting Reactant Problems 1.39 g NO 0.788g NH 3 consumed and g NH 3 remains 1.50 g1.85 gX g 68 g160 g120 g X = 1.39g X = 2.65g Y NH 3 Remains (1.50g – 0.788g = g ) Y = 0.788g The conversion of ammonia to nitrogen monoxide is 4 NH 3(g) + 5 O 2(g)  4 NO (g) + 6 H 2 O (g) a)How many grams of NO form when 1.50 g of NH 3 reacts with 1.85 g O 2 ? b)Which reactant is the limiting reactant and which is the excess reactant? c)How much of the excess reactant remains after the limiting reactant is completely consumed?