3Goals of Chapter 9Understand molecular and molar mass given in balanced equationUse balanced equation to determine the relationships between moles of reactants and moles of productsRelate masses of reactants and products in a chemical reactionPerform mass calculations that involve scientific notationUnderstand the concept of limiting reactantsRecognize the limiting reactant in a reactionUse the limiting reactant to do stoichiometric calculations
4Information Given by Chemical Equations Reactions are described by equationsGive the identities of the reactants & productsShow how much of each reactant and product participates in the reaction.Numbers or coefficients enable us to determine how much product we get from a given quantity of reactants
5Information Given by Chemical Equations Atoms are rearranged in a chemical reaction (not created or destroyed)Must have same number of each type of atom on both sides of equation.Coefficients give relative number of molecules, meaning we can multiply by any number and still have a balanced equation.
7C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g) Combustion of PropanePropane reacts with oxygen to produce heat and the products carbon dioxide and water:C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)
8Interpretation of Equation 1 molecule of C3H8 reacts with 5 molecules of O2 to give 3 molecules of CO2 plus 4 molecules of H2O1 mole of C3H8 reacts with 5 moles of O2 to give 3 moles of CO2 plus 4 moles of H2O
9Nuts & Bolts of Chemistry Activity 2 Nuts (N) react with 1 bolt (B) to form a nut- bolt molecule2N + B → N2BNote difference between coefficient and subscriptConstruct nut-bolt moleculesWhich is limiting reactant? Why?
10Average mass of bolt = 10. 64 g & average mass of nut = 4 Average mass of bolt = g & average mass of nut = 4.35 g; If you are given about 1500 g of each, answer the following questions: 1. How many bolts are in 1500 g? How many nuts are in 1500 g? 2. Which is limiting reactant? Why? 3. What is largest possible mass of product? How many products can you make? 4. What is mass of leftover reactant?
11Mole-Mole Relationships 2H2O(l) → 2H2(g) + O2(g)Equation tells us that 2 mol of H2O yields 2 mol of H2 and 1 mol of O2If we decompose 4 mol of water, how many moles of products do we get?4H2O(l) → 4H2(g) + 2O2(g)If we decompose 5.8 mol of water, how many moles of products do we get?5.8H2O(l) → 5.8H2(g) + ?O2(g)
12Mole Ratios: conversion factors based on balanced chemical equations From initial equation:2 mol H2O = 2 mol H2 = 1 mol O2Can use equivalent statement & perform dimensional analysis5.8 mol H2O x 1 mol O2_ = 2.9 mol O22 mol H2O
13Mass CalculationsWhat mass of oxygen (O2) is required to react with exactly 44.1 g of propane (C3H8)?
15Step 2: Convert grams of propane to moles of propane 44.1 g C3H8 x 1 mol C3H8 = 1.00 mol C3H844.09 g C3H8
161.00 mol C3H8 x 5 mol O2 = 5.00 mol O2 1 mol C3H8 Step 3: Use coefficients in equation to determine moles of oxygen required1.00 mol C3H8 x 5 mol O2 = 5.00 mol O21 mol C3H8
17Step 4: Use molar mass of O2 to calculate the grams of oxygen 5.00 mol O2 x g O2 = 160 g O21.00 mol O2
18Can perform conversion in on long step: 44.1 g C3H8 x 1 mol C3H x 5 mol O2 x g O2 = 160 g O244.09 g C3H mol C3H mol O2
19Mass Calculations Using Scientific Notation Step 1: Balance the equation for the reactionStep 2: Convert the masses of reactants or products to molesStep 3: Use the balanced equation to set up the appropriate mole ratio(s)Step 4: Use the mole ratio(s) to calculate the number of moles of the desired product or reactant.Step 5: Convert from moles back to mass
20StoichiometryThe process of using a chemical equation to calculate the relative masses of reactants and products involved in a reaction
21Mass Calculations: Comparing Two Reactions Antacids are used to neutralize excess hydrochloric acid secreted by the stomach. Which antacid is more effective: baking soda, NaHCO3, or milk of magnesia, Mg(OH)2?Determine how many moles of stomach acid (HCl) will react with 1.00 g of each acid.
26This reaction contained the exact number of molecules to make ammonia molecules with no unreacted molecules left over. Before the reaction, there were 15 H2 molecules and 5 N2 molecules which gives the exact ratio to make ammonia, 3:1. This type of mixture is called a stoichiometric mixture – contains the relative amounts of reactants that matches the numbers in balanced equation
27What happens when the ratio is not the same as in the chemical equation?
29Before and after the reaction. (Some N2 molecules are left over)
30H2 is limiting reactant in previous slide The reactant that runs out first and thus limits the amounts of products that can formH2 is limiting reactant in previous slide
31Ammonia, which is used as a fertilizer, is made by combining nitrogen from the air with hydrogen. The hydrogen is produced by reacting methane (CH4) with water. If you have 249 grams of methane, how much hydrogen will be produced and how much water will you need to convert all of the methane to hydrogen?
33Figure 9.1: A mixture of 5CH4 and 3H2O molecules.
34Step 2: Convert mass of methane to moles 249 g CH4 x 1 mol CH4 = 15.5 mol CH416.04 g CH4
35Step 3: Determine moles of H2O needed 15.5 mol CH4 x 1 mol H2O = 15.5 mol H2O1 mol CH4
36Step 4: Determine mass of water 15.5 mol H2O x g H2O = 279 g H2O1 mol H2O
37Reacting 279 grams of water with 249 grams of methane will cause both reactants to “run out” at the same time. If 300 grams of water is reacted with 249 grams of methane, the methane will “run out” first = limiting reactant (it limits the reaction)
38Figure 9. 2: A map of the procedure used in Example 9. 7 Figure 9.2: A map of the procedure used in Example 9.7. (see for determining limiting reactant)
39Steps for solving stoichiometric problems involving limiting reactants: Step 1: Write and balance chemical equationStep 2: Convert known masses to molesStep 3: Using number of moles of reactants and mole ratios, determine limiting reactantStep 4: Use amount of limiting reactant and mole ratios to calculate number of moles of productStep 5: Convert from moles to mass
40Theoretical YieldCalculated yield from chemical reaction (amount of product form mole ratio calculations)Maximum amount that can be producedAmount predicted is seldom obtainedSide reactions occurActual yield: amount of product actually obtained
41Actual Yield x 100% = percent Theoretical Yield yield Percent YieldActual Yield x 100% = percentTheoretical Yield yield