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I conducted an experiment. I did a reaction were I started with aluminum and remade aluminum crystals. I began with .5 g of Aluminum foil but when I weighed the collection (where I added ethanol) I weighed 6g. Now if the weight of the ethanol was 5.753g what is my actual and theoretical yields?

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Related questions

DATA FROM EXPERIMENT:

MASS OF WEIGHING PAPER = 0.3670 g
MASS OF WEIGHING PAPER + Al FOIL = 1.0000 g
MASS OF ALUM FOIL (NET WEIGHT OF ALUM) = 0.6330 g
MASS OF EMPTY 50 mL BEAKER = 29.3380 g
MASS OF BEAKER + ALUM CRYSTALS = 39.8840 g
MASS OF ALUM (NET WEIGHT OF ALUM) = 10.5460 g

we added about 35 mL of de-ionized water and 12mL of 6M KOHsolution to the beaker. I converted the molarity to moles and got6M KOH (.012 L) = .072 mol KOH

When metallic aluminum comes into contact with aqueous solutions ofstrong bases like potassium hydroxide, KOH, it reacts to formhydrogen gas and a salt containing both aluminum and potassiumions.

2 Al (s) + 2 KOH (aq) + 6 H2O (l) ? 2 KAl(OH)4 (aq) + 3 H2(g)

This salt, KAl(OH)4, is soluble in solutions of strong acids, suchas sulfuric acid, H2 SO4, and reacts to form aluminum hydroxide,Al(OH)3, and potassium sulfate, K2 SO4.

2 KAl(OH)4 (aq) + H2 SO4 (aq) ? 2 Al(OH)3 (s) + K2 SO4 (aq) + 2 H2O(l)

Excess sulfuric acid is required to dissolve the Al(OH)3, and toform aluminum sulfate, Al2 (SO4)3.

2 Al(OH)3 (s) + 3 H2 SO4 (aq) ? Al2 (SO4)3 (aq) + 6 H2O (l)

As this solution is cooled, the aluminum and potassium sulfatesalts crystallize out of the solution as the alum. The equationbelow shows only the ions and the water involved in thecrystallization process.

K+ (aq) + Al3+ (aq) + 2 SO4 2- (aq) + 12 H2O (l) ? KAl(SO4)2 * 12H2O (s)

MY QUESTION IS.. I DONT KNOW WHICH EQUATION TO USE TO CALCULATE THEPERCENT YIELD. SHOULD IT BE THE FIRST ONE SINCE IT IS BEFORE ANY OFTHE PROCEDURE WAS DONE? BUT FROM THE FIRST EQUATION, IT DID NOTPRODUCE THE ALUM CRYSTALS YET : KAl(SO4)2 * 12H20

EQUATION FOR PERCENT YIELD:
100 % x [mass of actual yield / mass of theoretical yield]

Hi, so I am unsure if this could be answered or not, but for my Organic Chemistry laboratory, I performed an aldol Condensation. The reaction used 0.25g of 2,4-dimethoxybenaldehyd and 0.20g of 1-indanone. I added 0.5g of NaOH. The reaction stood for 15 minutes and began to harden. I then added 2ml of HCL to the tested the basicity of the solution. I then vacuum filtrated the product and rinsed with HCl. Then I began the recrystallization of the product from 90% warm ethanol. The solution cooled and then was transferred to an ice bath. During this time, I was unable to obatain any crystals from my product. I am having trouble figuring out where I went wrong in my experiement to not get an crystals.

I appreciate any help/suggestions. Thank you in advance.

please i want the answer as soo as possible with steps

thank you

best

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