Presentation on theme: "Chemical Reactions Unit"— Presentation transcript:
1Chemical Reactions Unit Learning Goal 4: Use molar relationships in a balanced chemical reaction to predict the mass of product produced in a simple chemical reaction that goes to completion.
2The equation above gives the recipe to make one S’more. S’MoresSubstanceSymbolUnit MassGraham CrackerS7.0 gMarshmallowMm7.1 gChocolate PiecesOr3.3 gS’moreS2MmOr3__________ g2S + Mm + 3Or S2MmOr3The equation above gives the recipe to make one S’more.
4Calculating Masses of Reactants and Products Balance the equation.Convert mass to moles.Set up mole ratios.Use mole ratios to calculate moles of desired molecule.Convert moles to grams or to particles, if necessary.
5Working a Stoichiometry Problem 6.50 grams of aluminum reacts with an excess of oxygen. How many grams of aluminum oxide are formed.Identify reactants and products and write the balanced equation.4Al + 3O2 2Al2O3What are the reactants?Every reaction needs a yield sign!What are the products?What are the balanced coefficients?
6Working a Stoichiometry Problem 6.50 grams of aluminum reacts with an excessof oxygen. How many grams of aluminum oxideare formed?4 Al O2 2Al2O36.50 g Al1 mol Al2 mol Al2O3g Al2O3=? g Al2O326.98 g Al4 mol Al1 mol Al2O36.50 x 2 x ÷ ÷ 4 =12.3 g Al2O3
7Calculating Percent Composition of a Compound Like all percent problems:partwholeFind the mass of each of the components (the elements),Next, divide by the total mass of the compound; then x 100x 100 % = percent
8ExampleCalculate the percent composition of a compound that is made of 29.0 grams of Ag with 4.30 grams of S.29.0 g AgX 100 = 87.1 % Ag33.3 g totalTotal = 100 %4.30 g SX 100 = 12.9 % S33.3 g total
9Percent Yield Theoretical Yield Actual Yield Percent Yield The amount of product predicted from the amount of reactant used.The maximum amount of product that could form.Actual YieldThe amount of product that is actually produced.Percent YieldActual Yield X 100% = Percent Yield
10First write the balanced equation. 2H2 + CO CH3OH Suppose 68,500 g of CO is reacted with 8,600 g of H2 to produce CH3OH. Calculate the theoretical yield of methanol. If 35,700 of CH3OH is actually produced, what is the percent yield of methanol?First write the balanced equation.2H2 + CO CH3OHNext Calculate moles of reactants.68,500 g CO 1 Mol CO = 2450 mol CO28.01 g CO8600 g H Mol H = 4270 mol H22.016 g H2
11Cont. Next determine the limiting factor 2,450 mol CO 2 mol H2 = 4,900 mol H21 mol CO4270 mol H2 < 4,900 mol H2Hydrogen is the limiting factor.Next determine the maximum amount of methanol that can be produced in this reaction.4270 mol H2 1 mol CH3OH = mol CH3OH2 mol H2
12Cont. Convert moles to grams. 2140 mol CH3OH g CH3OH= g CH3OH1 mol CH3OHPercent Yield35700 g CH3OH x 100 % = 52.0%68600 g CH3OH
13Formulas molecular formula = (empirical formula)n [n = integer] molecular formula = C6H6 = (CH)6empirical formula = CHEmpirical formula: the lowest whole number ratio of atoms in a compound.Molecular formula: the true number of atoms of each element in the
14Limiting Reactant The limiting reactant is the reactant that is consumed first, limiting the amounts of products formed.
20Solving a Limiting Reactant Stoichiometry Problem Write and balance the equation for the reaction.Convert known masses of reactants to moles.Determine which reactant is limiting.Using the amount of the limiting reactant and the appropriate mole ratios, compute the number of moles of the desired product.Convert moles to mass (if required).
21Suppose 25. 0 kg of nitrogen gas and 5 Suppose 25.0 kg of nitrogen gas and 5.00 kg of hydrogen gas are mixed and reacted to form ammonia. Calculate the mass of ammonia produced when this reaction is run to completion.Write the balanced equation.N2 + 3H2 2NH3Convert known masses to moles.25000 g N2 1 mol N = mol N228.02 g N25000 g H2 1 mol H2 = 2,480 mol H22.016 g H2
22Cont. Determine the Limiting reactant. 892 mol N2 3 mol H2 = 2676 mol H21 mol N22480 mol H2 < 2676 mol H2Hydrogen is the limiting reactant. To get all of the Nitrogen to react you would need 2676 mol of hydrogen gas.
23Cont.Using the amount of the limiting reactant and the appropriate mole ratios, compute the number of moles of the desired product.2480 mol H2 2 mol NH3 = mol NH33 mol H2
24Cont. Convert moles to mass. Calculate the molar mass of NH3 N x 1 = 14.01H X 3 =17.0341653 mol NH g NH3 = 28,150 g NH mol NH3