Chem 1075 Chapter 10 Stoichiometry Lecture Notes

Transcription

1 Chem 1075 Chapter 10 Stoichiometry Lecture Notes Slide 2 What is stoichiometry? Chemists and chemical engineers must perform calculations based on balanced chemical reactions to predict the cost of processes. These calculations are used to avoid using large excess amounts of costly chemicals. The calculations these scientists use are called stoichiometry calculations. Slide 3 Interpreting Chemical Equations Slide 4 Moles and Equation Coefficients NO (g) O2(g) NO2(g) molecules molecule molecules molecules molecules molecules molecules molecules molecules moles mole moles Slide 5 Mole Ratios (Coefficient Ratios) We can now read the balanced chemical equation as gas react with gas to produce gas. The coefficients indicate the, or the ratio of the moles, of reactants and products in every balanced chemical equation. This mole ratio is also called the

2 Slide 6 Interpretation of Coefficients From a balanced chemical equation, we know how many of a substance react and how many of product(s) are produced. If there are gases, we know how many of gas react or are produced. Slide 7 Conservation of Mass The law of conservation of mass states that mass is neither created nor destroyed during a chemical reaction. Lets test: 2 NO(g) + O 2 (g) 2 NO 2 (g) Slide 8 Mole Mole Relationships We can relate to in a reaction by using coefficients (CR = coefficient relationship) from the balanced equation. Slide 9 Mole Mole Relationships Use the balanced chemical equation to write CR which can be used as unit factors: N 2 (g) + O 2 (g) 2 NO(g) Since 1 mol of N 2 reacts with 1 mol of O 2 to produce 2 mol of NO, we can write the following mole relationships: Slide 10 Mole Mole Calculations How many moles of NO will be formed when 2.25 mol of nitrogen react? N 2 (g) + O 2 (g) 2 NO(g)

3 Slide 11 Mass Mass Problems In a mass-mass stoichiometry problem, we will convert a of a reactant or product to an of reactant or product. There are three steps: Convert the given mass to moles using the as a unit factor. Convert the moles of given to moles of the unknown using the in the balanced equation. Convert the moles of unknown to grams using the as a unit factor. Slide 12 Mass Mass Problems Slide Mass Mass Stoichiometry Problem What is the mass of mercury produced from the decomposition of 1.25 g of orange mercury (II) oxide? 2 HgO(s) 2 Hg(l) + O 2 (g) Slide 15 Limiting Reactant Concept Say you re making grilled cheese sandwiches. You need 1 slice of cheese and 2 slices of bread to make one sandwich. 1 Cheese + 2 Bread 1 Sandwich If you have 5 slices of cheese and 8 slices of bread, how many sandwiches can you make? Slide 16 Limiting Reactant

4 Since you run out of bread first, bread is the ingredient that how many sandwiches you can make. In a chemical reaction, the is the reactant that controls the amount of products you can make. A limiting reactant is before the other reactants. The other reactants are present in excess and are referred to as. Slide Determining the Limiting Reactant If you heat 2.50 mol of Fe and 3.00 mol of S, how many moles of FeS are formed? Fe(s) + S(s) FeS(s) Slide 19 Mass Limiting Reactant Problems There are three steps to a limiting reactant problem: 1. Calculate the that can be produced from the reactant. mass reactant #1 mol reactant #1 mol product mass product 2. Calculate the that can be produced from the reactant. mass reactant #2 mol reactant #2 mol product mass product 3. The limiting reactant is the reactant that produces the amount of product. Slide Mass Limiting Reactant Problem How much molten iron is formed from the reaction of 25.0 g FeO and 25.0 g Al? 3 FeO(l) + 2 Al(l) 3 Fe(l) + Al 2 O 3 (s)

5 Slide 23 Percent Yield When you perform a, the amount of product collected is the. The amount of product calculated from a limiting reactant problem is the. The is the amount of the actual yield compared to the theoretical yield. Slide 24 Calculating Percent Yield Suppose a student performs a reaction and obtains g of CuCO 3 and the theoretical yield is g. What is the percent yield? Cu(NO 3 ) 2 (aq) + Na 2 CO 3 (aq) CuCO 3 (s) + 2 NaNO 3 (aq)