Slide 1Slide 2Percent Yields from ReactionsSlide 4Clicker questionSlide 6Slide 7Slide 8Concentration of SolutionsSlide 10Slide 11Slide 12Slide 13Slide 14Concentrations of SolutionsConcentrations of SolutionsClicker questionSlide 181What is the maximum mass of sulfur dioxide that can be produced by the reaction of 95.6 g of carbon disulfide with 110. g of oxygen? Carbon disulfide reacts with oxygen to form carbon dioxide and sulfur dioxide.CS2 + 3O2  CO2 + 2SO295.6 g CS2 / 76.13 g/mol = =1.26 mole CS2 (2 mole SO2 / 1 CS2)=2.52 mole SO2 (64.065 g/mole) =161 g SO2110 g O2 / 32 g/mol O2 = 3.4374 mole O2=3.4374 mole O2 (2 mole SO2 / 3 mole O2)=2.29 mole SO2 (64.065 g/mole)=147 g SO2How many g of silver bromide can be formed when solutions containing 50.0 g of magnesium bromide and 100.0 g of silver nitrate are mixed together (magnesium nitrate is also produced)? How many g of the excess reactant remain unreacted?MgBr2 + 2AgNO3  2AgBr + Mg(NO3)2***CONVERT GRAMS TO MOLESSSSSSSSS50.0 g MgBr2 / 184.1 g/mol = 0.272 mole MgBr2= 0.272 mole MgBr2 (2 AgBr / 1 MgBr2)= .544 mole AgBr **Limiting reactant=.544 (187.8 g/mol) = 102 g AgBr100.0 g AgNO3 / 196.88 g/ mol =0.589 mole AgNo3=0.589 mole AgNO3 (2 AgBr / 2AgNO3)=0.589 mole AgBr50.0 g MgBr / 184.1 g/mol = .272 mol MgBr=.272 mol MgBr (2 mol AgNO3 / 1 mol MgBr)=.544 mol AgNO3=.544 mol AgNO3 (169.9 g/mol) =92.48 g AgNO3100 g AgNO3 – 92.48 g AgNO3 = 7.6 g AgNO33Percent Yields from Reactions•Theoretical yield is calculated by assuming that the reaction goes to completion.–Determined from the limiting reactant calc.•Actual yield is the amount of a specified pure product made in a given reaction.–In the laboratory, this is the amount of product that is formed in your beaker, after it is purified and dried.•Percent yield indicates how much of the product is obtained from a reaction.% yield = actual yieldtheoretical yield100%4A 10.0 g sample of ethanol, C2H5OH, was boiled with excess acetic acid, CH3COOH, to produce 14.8 g of ethyl acetate, CH3COOC2H5. What is the percent yield?3 2 5 3 2 5 2CH COOH + C H OH CH COOC H H O� +Clicker questionIn a previous problem we determined that 102. g of AgBr was theoretically possible. But when we did the actual experiment we only made 85.6 g of AgBr. What is the % yield.85.6/102 x 100 = 83.9 %Sequential ReactionsMore than one reaction is required to convert starting materials into the desired products. The amount of desired product from each reaction is taken as the starting material for the next reaction.7Starting with 10.0 g of benzene (C6H6), calculate the theoretical yield of nitrobenzene (C6H5NO2) and of aniline (C6H5NH2).10.0 g C6H6 / 78.0 g/mol = .128 mol C6H6 (1 C6H5NO2 / 1 C6H6) = .128 MOL C6H5NO2 (123 G/MOL) = 15.8 G C6H5NO2.128 MOL C6H5NO2 (1 C6H5NH2 / 1 C6H5NO2) = .128 MOLE C6H5NH2 (93.0 G/MOL) = 11.9 G C6H5NH2Clicker questionIf 6.7 g of aniline is prepared from 10.0 g of benzene, what is the percentage yield?6.7 / 11.9 X 100 = 56%9Concentration of Solutions•Solution is a mixture of two or more substances dissolved in another.–Solute is the substance present in the smaller amount.–Solvent is the substance present in the larger amount.–In aqueous solutions, the solvent is water.•The concentration of a solution defines the amount of solute dissolved in the solvent.–The amount of sugar in sweet tea can be defined by its concentration.10One common unit of concentration is:% by mass of solute has the symbol % w/wmass of solute% by mass of solute = 100%mass of solutionmass of solution = mass of solute + mass of solvent�11What mass of NaOH is required to prepare 250.0 g of solution that is 8.00% w/w NaOH?X g NaOH / 250.0 g (8.00 g/100g) x = 20.0 g NaOH12Calculate the mass of 8.00% w/w NaOH solution that contains 32.0 g of NaOH.32.0 g / x g (8/100)x = 400 g13Calculate the mass of NaOH in 300.0 mL of an 8.00% w/w NaOH solution. Density is 1.09 g/mL.14What volume of 12.0% w/w KOH contains 40.0 g of KOH? The density of the solution is 1.11 g/mL.15Concentrations of SolutionsSecond common unit of concentration:mLmmolLmolessolution of liters ofnumber solute of moles ofnumber molarity MM16Concentrations of SolutionsCalculate the molarity of a solution that contains 12.5 g of sulfuric acid in 1.75 L of solution.Clicker questionWhat is the molarity of a solution that contain 10.0 moles of a solute in 2.00 l of solution?18Determine the mass of calcium nitrate required to prepare 3.50 L of a 0.800 M Ca(NO3)2