A 10.0 g sample of ethanol, C2H5OH, was boiled

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A 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?

Ch3 #52 Sequential Reactions CH3CH2Cl + Mg CH3CH2MgCl CH3CH2MgCl + H2O CH3CH3 + Mg(OH)Cl The Grignard reaction is a two-step reaction used to prepare pure hydrocarbons. We allow 27.2g of CH3CH2Cl to react with excess magnesium. CH3CH2MgCl is obtained in 79.5% yield. In the second step a 85.7% yield of CH3CH3 is obtained. What mass of CH3CH3 is collected? 1 mol CH3CH2Cl 64.4 g CH3CH2Cl 1 mol CH3CH2Cl 27.2g CH3CH2Cl * * = 0.4224mol 0.4224 mol CH3CH2MgCl*0.795=0.3358 mol CH3CH2MgCl 1 mol CH3CH3 1 mol CH3CH2MgCl 0.3358mol CH3CH2MgCl* *0.857=0.2878mol CH3CH3 30.0g CH3CH3 1 mol CH3CH3 0.2878 mol CH3CH3* = 8.63 g CH3CH3

Solutions Parts of A (by mass) 100 parts mixture (by mass) % Composition %A (mass)= grams of Solute 100 g of Solution % Solute= Mass of Solution = Mass of Solute + Mass of Solvent

What mass of NaOH is required to prepare 250.0 g of solution that is 8.00% w/w NaOH? 8.00 g NaOH 100 g solution * 250.0 g solution = 20.0 g NaOH Calculate the mass of 8.00% w/w NaOH solution that contains 32.0 g of NaOH 100 g solution 8.00 g NaOH * 32.0g NaOH= 400.0 g Solution

Calculate the molarity of a solution that contains moles of solute L of solution Molarity= = M Calculate the molarity of a solution that contains 12.5 g of sulfuric acid in 1.75 L of solution. 12.5 g H2SO4 1.75 L solution 1 mole H2SO4 98.1 g H2SO4 * = 0.0728M

Determine the mass of calcium nitrate required to prepare 3.50 L of 0.800 M Ca(NO3)2

The specific gravity of concentrated HCl is 1.185 and it is 36.31% HCl. What is its molarity? 36.31 g HCl 100 g solution Sp. Gr. = 1.185  1.185 g soln ml of soln

Dilution M1V1=M2V2 What is the molarity of a solution that is prepared with 2.00ml of 4.00*10-2M K2CrO4 and brought up to a volume of 500ml? M1= 4.00*10-2M V1=2.00ml M2=? V2=500ml 4.00*10-2M*2.00ml=XM*500ml 4.00*10-2M*2.00ml 500 ml X= = 1.60*10-4M

What volume of 0.500 M BaCl2 is required to completely react with 4.32 g of Na2SO4? mol Na2SO4 142 g Na2SO4 mol BaCl2 mol Na2SO4 1 L solution 0.500 moles BaCl2 4.32 g Na2SO4 * * * =0.0608 L of 0.500M BaCl2 solution = 60.8ml of 0.500M BaCl2 solution