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Week 3 notes

by: Moira Notetaker

Week 3 notes CHM2045

Moira Notetaker
Chemistry 1
George Gower

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About this Document

These notes cover what we went over in this past week in lecture. Since we did not have class on Monday, we only had lecture Wednesday and Thursday!
Chemistry 1
George Gower
Class Notes
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This 5 page Class Notes was uploaded by Moira Notetaker on Sunday January 24, 2016. The Class Notes belongs to CHM2045 at University of Florida taught by George Gower in Fall 2016. Since its upload, it has received 28 views. For similar materials see Chemistry 1 in Chemistry at University of Florida.


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Date Created: 01/24/16
CHM2045 Gower 12116 Two Equation Problem What maximum number of moles of D can be prepared if 200 mol of A are combined with 200 mol of B and equation 1 is 75 ef cient and equation 2 is 50 ef cient Equation 1 2A B 4c Equation 2 3c 20 E We must go through several steps using the given values before we can actually gure out how much D can be prepared Start by guring out if A or B is the limiting reactant in the reaction using the given values 0 2 mole4 mol C2 mol A 4 mol C o 2moBx4moC1mol B8molC A is the limiting reactant and B is the excess reactant This simpli es the problem a bit as equation 1 already begins with 2 mol of A and we can tell that 2 mol ofA will make 4 mol of C However you must remember we are told equation 1 is only 75 ef cient so the actual yield of equation 1 will be 3 mol C 4 mol C x 075 3 mol C Now we can use the amount of C to nd the maximum amount of D we can have prepared 0 3mole2 molD3 molC2molD This works out well like with equation 1 and we know that 2 mol of D can be made with 3 mol of C However we must remember that we are told that equation 2 is only 50 ef cient so we must take that into account before we submit a nal answer 0 2moDx050 1moD The nal answer will be 1 mol of D can be prepared from 2 mol ofA Solution Stoichiometry Concentration of a solution can be found by using amount of solute present in the given quantity of the solvent or solution 0 M molarity moles of soluteliters of solution x 100 molL What mass of Kl is needed to make 500mL of a 28 M Kl solution Begin by converting mL to L o 500mL 0500L Set unknown number of moles X 28 M X mol0500L 28 M x 0500L X mol 14 X mol You now know the moles of Kl needed to make 0500L of a 28 M solution and you can use stoichiometry to convert to grams and get the mass of Kl needed o 14 mol Kl x 166g Kl1 mol Kl 232g of Kl Dissolving Solutions 0 One mole of an ionic solidnonmeta metal compound yields two or more moles of ions when dissolved in water 0 Solution are electrolytic o KBrs KaqBr39aq 1 mole of ionic sold 1 mole of cation 1 mole of anion 2 moles of product 0 Na2CO3s 2Naaq CO3239aq 1 mole of ionic solid 2 moles of cation 1 mole of anion 3 moles of product If you dissolve 5 mol of NH42SO4 in H20 how many mol of NH4 and 504239 are in the resulting solution 0 Try writing this equation yourself the answer is 10 mol NH4 and 5 mol 504239 What is the mass you would need to prepare 250mL of a 030 M K2Cr04 solution 030M x X mol0250L 0075 mol K2Cr04 x 1942g1 mol 15g K2Cr04 75mL of a 20 M solution is combined with 125mL of a 40 M Na2504 What is the resulting molarity of the Na ions in the mixture 0 This important to understand so I will walk you through it but you must do the math yourself Begin by nding the amount of moles of the solvent in each solution Convert from moles of solution to moles of Na for each amount of moles Add together the moles of Na and divide sum by 0200L125mL 75mL You should end up with 58 M of Na Dilution 0 Procedure for preparing less concentrated solution from a more concentrated solution 0 REMEMBER this equation MiVi Mfo M initial molarity Vi initial volume Mf nal molarity Vf nal volume 0000 CHM2045 Gower 12016 How To Read Chemical Equations 2Mg02 2Mg0 o 2 moles of magnesium 1 mole of oxygen becomes 2 moles of magnesium oxide 0 REMEMBER mass of the reactants must equal the mass of the products 0 The chemical equation must be balanced Meaning you must have the same amount of moles of an element on the reactant side as you do on the product side 0 Hint combustion reactions like this one usually create a product that ends in quotoxidequot Try to balance C2H6 02 2co2 3H20 Note rst that the only element needing to be balanced is oxygen as carbon and hydrogen already have been balance on both sides of the equation 0 Count the number of oxygen moles on the product side of the equation 2 x 023 x 0 7 oxygen moles on product side 0 C2H6 72o2 2co2 3H20 Simply by multiplying all coef cients by 2 o 2C2H6 702 4co2 6H20 0 That is your balanced equation with 4 moles of carbon on both sides 12 moles of hydrogen on both sides and 14 moles of oxygen on both sides Methanol burns in air If 209 g of methanol are used in the reaction what mass of H20 is produced 0 Begin with the balanced equation 2CH30H 302 2C02 4H20 You must use stoichiometry to go from grams of methanol to grams of water 209g CH30H x 1 mol CH30H32042g CH30H x 4 mol H202 mol CH30H x 18016g H201 mol H20 235g of H20 REMEMBER When doing these kinds of problems you are operating under the assumption that there is enough reactant to get a full productthat there is enough 02 for all the methanol to react fully Limiting Reactants You have a limiting reactant when there is not enough of one reactant to let all of the other reactant get fully used up in the reaction 0 Usually this is a reactant that needs a coef cient in front of it to balance the equa on o 2N002 2N02 o In this case we needed 2 moles of NO to react with 1 mole of 02 therefore N0 is the limiting reactant and 02 is the excess reactant 124g of Al reacted with 601g of Fe203 What is the mass of the product Al203 Begin by writing out the balanced equation to get the molar ratios 0 F8203 A203 2Fe Use stoichiometry to nd the moles of aluminum oxide created when 124g Al reacts fully and when 601g of Fe203 reacts fully 124g Al x 1 mol Al2698g Al x 1 mol Al2032 mol Al 23 mol Al203 601g Fe203x 1 mol Fe2031597g Fe203 x 1 mol Al2031 mol Fe203 376 mol Al203 Based on the above equations we can tell that Al is the limiting reactant in this problem and iron III oxide was the excess reactant We also now aware that the most Al203 that can be produced is 23 mol which converts to 235g of Al203 Note to nd amount of excess reactant is leftover just convert from moles of limiting reactant to moles of excess reactant and subtract that from the moles of excess reactant available 0 124g Al x 1 mol Al2698g Al x 1 mol Fe2032 mol Al 23 mol Fe203 o 376 mol Fe203 23 mol Fe203 15 mol Fe203 leftover Reaction Yield 0 Theoretical Yield what you calculate to be the mass of the product in a reaction 0 Actual Yield the actual amount of product created by the reaction 0 Percent Yield Actual yieldTheoretical yield x 100 Double Unknown questions are the hardest you will encounter on this exam A 2920g mixture of benzene C6H6 and toluene C7H8 is combusted in excess 02 and 9809g of C02 is yielded What were the individual masses of benzene and toluene in the original mixture 0 Begin by converting carbon dioxide to moles o 9809g C02 x 1 mol C024401g C02 223 mol C02 0 You know from 1 mole of C6H6 you will get at least 6 moles of C02 to balance the equation 0 For the same reason you know you will have at least 7 moles of CO2 for each mole of C7H8 Set x grams of benzene Set moles of C02 produced to the stoichiometry you will need to get grams of C6H6 and C7H8 O g C6H6 X mol C6H6 X mol mol C6H6 Use algebra to solve for x and you will have your grams of benzene and toluene 0 12059 benzene and 17159 of toluene


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