Chapter 4 Aqeous RXN

Chapter 4 Aqeous RXN - Determining Limiting Reagents Guided...

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Determining Limiting Reagents Guided Practice Problem Part of the SO 2 that is introduced into the atmosphere ends up being converted to sulfuric acid, H 2 SO 4 . The net reaction is: 2SO 2 (g) + O 2 (g) + 2H 2 O( l ) 2H 2 SO 4 ( aq ) How much H 2 SO 4 can be formed from 5.0 mol of SO 2 , 1.0 mol O 2 , and an unlimited quantity of H 2 O?
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Consider the following reaction: 2Na 3 PO 4 ( aq ) + 3Ba(NO 3 ) 2 ( aq ) Ba 3 (PO 4 ) 2 (s) + 6NaNO 3 ( aq ) Suppose that a solution containing 3.50 g of Na 3 PO 4 is mixed with a solution containing 6.40 g of Ba(NO 3 ) 2 . How many grams of Ba 3 (PO 4 ) 2 can be formed? What is the % yield, if experimentally , only 4.70 g were obtained from the reaction? Determining Limiting Reagents Guided Practice Problem
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Chapter 4 Chapter 4 Aqueous Reactions and Solution Aqueous Reactions and Solution Stoichiometry Stoichiometry CHEMISTRY The Central Science 9th Edition
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Solutions are homogenous mixtures of two or more substances: Solute: present in smallest amount and is the substance dissolved in the solvent. Solvent: present in the greater quantities and is used to dissolve the solute. Example: NaCl dissolved in Water (water = Solvent and NaCl = Solute) Change concentration by using different amounts of solute and solvent. Molarity: Moles of solute per liter of solution. Solution Composition Solution Composition
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Formula for Molarity The most widely used way of quantifying concentration of solutions in chemistry. Molarity is generally represented by the symbol M and defined as the number of moles of solute dissolved in a liter of solution . liters in solution of volume solute of moles Molarity = Concentrations of Solutions Concentrations of Solutions
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Calculate the molarity of a solution made by dissolving 23.4 g of sodium sulfate, Na 2 SO 4 , in enough water to form 125 mL of solution. Class Guided Practice Problem Class Guided Practice Problem Class Practice Problem Calculate the molarity of a solution made by dissolving 5.00 g of NaCl in sufficient water to form 0.125 L of solution.
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Key concept: If we know: molarity and liters of solution, we can calculate moles (and mass) of solute. Class Guided Practice Problem Class Guided Practice Problem Determining Mass using Molarity How many grams of C 6 H 12 O 6 are required to make 100 mL of 0.278 M C 6 H 12 O 6 ? Class Practice Problem How many grams of NaCl are required to make a 1 L of 0.500 M NaCl?
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Solutions are routinely prepared in stock solutions form. Example: 12 M HCl Solution of lower concentrations are prepared by adding more solvent (e.g., water), a process called dilution . We recognize that the number of moles are the same in dilute and concentrated solutions. Hence, moles solute before dilution = moles solute after dilution
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This note was uploaded on 02/20/2011 for the course CHEM 101 taught by Professor John during the Spring '11 term at Caldwell College.

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Chapter 4 Aqeous RXN - Determining Limiting Reagents Guided...

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