3Chemical EquationsConcise representations of chemical reactions
4Anatomy of a Chemical Equation CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)
5Anatomy of a Chemical Equation CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)Reactants appear on the left side of the equation.
6Anatomy of a Chemical Equation CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)Products appear on the right side of the equation.
7Anatomy of a Chemical Equation CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)The states of the reactants and products are written in parentheses to the right of each compound.
8Anatomy of a Chemical Equation CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)Coefficients are inserted to balance the equation.
9Subscripts and Coefficients Give Different Information Subscripts tell the number of atoms of each element in a molecule
10Subscripts and Coefficients Give Different Information Subscripts tell the number of atoms of each element in a moleculeCoefficients tell the number of molecules
11Stoichiometric Calculations The coefficients in the balanced equation give the ratio of moles of reactants and products
12Hints for BalancingIf an element occurs in only one compound on each side of the equation, balance that firstBalance free elements lastBalance polyatomic ions as a group unit if they remain unchangedIf you get stuck, try doubling everything!
14Example 4-1 A Balance the equations: H3PO4 + CaO Ca3(PO4)2 + H2O C3H8 + O2 CO2 + H2O
15Example 4-2AWrite a balanced equation to represent the reaction of mercury(II) sulfide and calcium oxide to produce calcium sulfide, calcium sulfate and mercury metal.
16(s) Solid (l) Liquid (g) Gas (aq) aqueous Over the arrow: catalyst, heat or conditions needed
174.2 Chemical Equations and Stoichiometry To Measure Elements!
18Stoichiometric Calculations From the mass of Substance A you can use the ratio of the coefficients of A and B to calculate the mass of Substance B formed (if it’s a product) or used (if it’s a reactant)
19Stoichiometric Calculations C6H12O6 + 6 O2 6 CO2 + 6 H2OStarting with 1.00 g of C6H12O6…we calculate the moles of C6H12O6…use the coefficients to find the moles of H2O…and then turn the moles of water to grams
20Switching SubstancesYOU CAN ONLY SWITCH SUBSTANCES IN MOLES!! Must use balanced equation!
21Example 4-3AHow many moles of O2 are produced from the decomposition of 1.76 moles of potassium chlorate?2 KClO3(s) 2 KCl(s) + 3O2(g)
22Example 4-4AHow many grams of magnesium nitride are produced by the reaction of 3.82 grams of Mg and excess N2?
23Example 4-5AFor the reaction NH3 + O2 N2 + H2O How many grams of NH3 are consumed per gram of O2?
30Example 4-10AA 15.00mL sample of 0.450M K2CrO4 is diluted to mL. What is the concentration of the new solution?
31Example 4-11AHow many milliliters of M K2CrO4 must be added to excess AgNO3(aq) to produce 1.50 g Ag2CrO4?
324.4 Determining the Limiting Reactant Limiting Reactants
33How Many Cookies Can I Make? You can make cookies until you run out of one of the ingredientsOnce this family runs out of sugar, they will stop making cookies (at least any cookies you would want to eat)
34How Many Cookies Can I Make? In this example the sugar would be the limiting reactant, because it will limit the amount of cookies you can make
35Limiting ReactantsThe limiting reactant is the reactant present in the smallest stoichiometric amount
36Limiting ReactantsThe limiting reactant is the reactant present in the smallest stoichiometric amountIn other words, it’s the reactant you’ll run out of first (in this case, the H2)
37Limiting ReactantsIn the example below, the O2 would be the excess reagent
38Example 4-12AIf 215 g P4 is allowed to react with 725 g Cl2 how many grams of PCl3 are formed?P4 + 6Cl2 4PCl3
39Example 13AFrom 12A, which reactant is in excess and what mass of the reactant remains after the reaction to produce PCl3?
404.5 Other Practical Matters in Reaction Stoichiometry
41Theoretical YieldThe theoretical yield is the amount of product that can be madeIn other words it’s the amount of product possible as calculated through the stoichiometry problemThis is different from the actual yield, the amount one actually produces and measures
42Percent Yield Actual Yield Theoretical Yield Percent Yield = x 100 A comparison of the amount actually obtained to the amount it was possible to makeActual YieldTheoretical YieldPercent Yield = x 100
43Example I2 is limiting and 2.57 g is the theoretical yield 1.20 g Al reacts with 2.40 g I2. What is the limiting reactant?2Al + 3 I2 2AlI31st find the reactant that produces the smallest amount of product.1.20 g Al x 1mol Al x 2 mol AlI3 x g AlI3 = 18.1 g AlI327 g Al mol Al mol AlI32.40 g I2 x 1 Mol I2 x 2 mol AlI3 x g AlI3 = 2.57 g AlI3253.8 g I mol I mol AlI3I2 is limiting and 2.57 g is the theoretical yield
44Theoretical Yield % Yield= actual Yield/ theoretical yield x 100 Say the actual yield was 2.05 gThen