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Chem 111 Week 4 notes

by: Notetaker

Chem 111 Week 4 notes 111/40551

Marketplace > University of St. Thomas > 111/40551 > Chem 111 Week 4 notes
User_39347_profile9613 Notetaker
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General Chemistry I

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Chapter 3 stoichiometry
General Chemistry I
Class Notes
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This 5 page Class Notes was uploaded by Notetaker on Sunday October 4, 2015. The Class Notes belongs to 111/40551 at University of St. Thomas taught by Uzcategui-White in Summer 2015. Since its upload, it has received 34 views.


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Date Created: 10/04/15
Chemistry Week 4 Lecture Notes 0 Limiting Reagent or reactant The reactant present in limited supply that determines the amount of a product formed The reactant that is limiting is consumed completely while some in excess remains unused Example 4 Al 5 3 02 9 2 A203 4 Al is limiting reactant solving problem limiting reactant produces the least amount of product 348 Many metals react with oxygen gas to form metal oxide For example calcium reacts as followed 2Ca s 02 g 9 2CaO s calculate the mass of calcium oxide that can be prepared from 420 g Ca and 280 g of 02 a what amount mol of Ca can be produced from the given mass of Ca mols grams molar mass 420 g Ca 4008 g mol 0015 mols Ca mols CaO 105 mols Ca x 2 mol CaO 2 mol Ca 105 mol CaO b What amount mol of CaO can be produced from given amount of 02 Mols 02 280 g 02 3200 mol 02 0875 mol 02 Mols CaO 0875 mol 02 x 2 CaO 1 mol 02 175 mol CaO c Which is the limiting reactant Calcium d How many grams of CaO can be produced Use that is from limiting reactant 105 mol CaO x 5808 g 1 gmol 589 g CaO e How many grams of the excess reactant remain Grams 02 excess Grams 02 initial 280 g Grams 02 reacted 9 use limiting reactant info Cannot use information from b because that is assuming all of the 02 reacted Mol 02 105 mol Ca x 1 mol 022 mol Ca 0525 mol 02 reacted Grams 02 0525 mol 02 x 3200 g 02 mol 02 168 g 02 consumed Initial amount 02 02 consumed answer 280 g 02 168g 02 112 g 02 o TheoreticalActualPercent Yield 0 Theoretical yield maximum mass of product that can be obtained from a reaction 0 Actual Yield mass of material that is actually obtained in the reaction 0 Percent yield actual mass of material maximum mass x 100 338 Percent yield of a reaction where 455 g Tungsten VI oxide W03 reacts with excess hydrogen gas to produce metallic tungsten and 960 mL of water W03 3H2 9 W 3H20 Grams H20 960 mL x 100 g mL 960 g H20 Mols H20 960 g H20 1800 gmol 533 mol H20 Mols W03 533 mol H20 x 1 mol W03 3 mol H20 178 mols W03 Mols W03 178 mols W03 x 2319 g W03 mol 412 g W03 reacted yield 412 g 455 g 905 Determining the Formula of an Unknown Compound a Empirical formula only gives the smallest whole number ratio of each type of atom b Molecular formula shows the actual number of atoms of each element in a molecule The molecular formula is a whole number multiple of the empirical formula c Structural formula 329 Find the empirical formula a 063 mol of chlorine atoms 22 mol oxygen atoms b 273 mass C 727 mass 02 a CL 063 063 0 22 063 Cl0 35 CL2 07 b 273 C 727 0 9 mols 273 g C 100 g compound x 100 gram compound1 273 g C mols C 273 g 1201 gmol 228 mol C mols 0 727 g 0 1600 gmol 454 mol 0 C 228228 0 454228 C02 33 cortisol 36247 gmol 696 C 834 H 221 0 Find the molecular formula Empirical x by whole number 9 molecular Mols C 696 g C1201 gmol 580 Mos H 834 g H 1008 g mol 834 Mos O 221 g 0 1600 gmol 138 C 580138 H 834138 O 138138 C4H60 empirical Molar mass empirical formula mass 36247 g mol 7004 gmol 518 5 C20 H30 05 Section 35 Fundamentals of solution stoichiometry A way of finding moles when you are working with solutions Concentration quantity of solute in a fixed amount of solvent Molarity M mols of solute volume of solution L Moles M xV L Dilution 9 of moles are the same 369 a volume mL of 226 M potassium hydroxide that contains 842 g of solute mos 842 g 561 gmol 150 mol KOH VL 150 mol 226 mL 0664 Lx 1000 mL 1L 664 mL 376 How many grams of solid barium sulfate form when 350 mL of 160 M barium chloride reacts with 580 mL of 065 M sodium sulfate Aqueous sodium chloride also forms Formula BaC2 aq Na2504 aq 9 BaSO4 s 2 NaCI aq Mos BaC2 160 mols volume x 350 x 10quot3 mL 0056 mols BaC2 Mols NaZSO4 065 M molsL x 580 x 10quot3 L 0038 mol NaZSO4 0038 mols NaZSO4 x 1 mol BaSO4 1 mol NaZSO4 x 23337 g BaSO4 1 mol BaSO4 87 g BaSO4 0056 mols BaCIZ x 1 mol BaSO4 1 mol BaCIZ x 23337 g BaSO4 1 mol BaSO4 131 g BaSO4


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