HomeClass Notes1,200,000US530,000

CHEM 1C Lecture Notes - Lecture 15: Aluminium Hydroxide, Barium Sulfate, Chemical Formula

22 views2 pages

pucesquirrel907

12 Apr 2020

School

Department

Chemistry and Biochemistry

Course

Professor

For unlimited access to Class Notes, a Class+ subscription is required.

Document Summary

In this lab, aluminum foil was used to synthesize alum, an inorganic salt with chemical formula kal(so4)2 12h2o. Various tests were performed to observe and confirm the presences of k+, al3+, so4. To synthesize the alum, 0. 256 grams of aluminum foil was weighed out, then 12. 5 ml of 4 m koh was added to produce the reaction. The addition of potassium hydroxide caused a vigorous fizzing and bubbling, as well as the formation of white foam and a minimal amount of condensation. The hydrogen gas was observed rising up from the test tube. This solution was filtered, and 10 ml of 9 m. H2so4 was added, which formed a solid aluminum hydroxide precipitate. This precipitate vanished when the solution was heated: After the solution was heated, the flask was placed in an ice bath to allow the alum crystals to form. The resulting crystals had a mass of 9. 891 grams, giving a percent yield of 122%.

Related textbook solutions

Chemistry: Structure and Properties

ISBN: 9780134293936

Basic Chemistry

ISBN: 9780134138046

Principles of Chemistry Molecular Approach

ISBN: 9780134112831

Principles of Chemistry Molecular Approach

3rd Edition, 2014

ISBN: 9780321971944

Chemistry: Structure and Properties

ISBN: 9780134293936

Chemistry: A Molecular Approach

ISBN: 9780321809247

Chemistry: A Molecular Approach

ISBN: 9780134874371

Principles of Chemistry: A Molecular Approach

ISBN: 9780134895741

Chemistry: The Central Science

14th Edition, 2017

ISBN: 9780134414232

Related Documents

CHEM 1C Lecture Notes - Lecture 17: Ammonium Hydroxide, Reagent, Crystal Field Theory

pucesquirrel907

CHEM 1C Lecture Notes - Lecture 18: Acetic Anhydride, Spearmint, Aspirin

pucesquirrel907

CHEM 1212 Chapter : Exp2 Synthesis Of Potassium Alum

azuremammoth471

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]

coffeeturtle150

Experiment 37 Qualitative Analysis of Group II Cations Precipitation and Separation of Group II lons The sulfides of the four Group II ions, Bi, Sn Sb, and Cu'", are insoluble at a pH of 0.5. The solution is adjusted to this plt and then saturated with H,S, which precipitates Bi,S, Sns, Sb,S, and Cus. The reaction with Bi" is typical: 2 Bis+ (aq) + 3 H,s(aq) ? Bizs,(s) + 6 H+ (aq) black Saturation with H,S could be achieved by simply bubbling the gas from a generator through the solution. A more convenicnt method, however, is to heat the acid solution after adding a small amount of thioacetamide This compound, CHCSNH;, hydrolyzes when heated in aqueous solution to liberate H,S CH,CSNH,(aq) + 2 H,01)? H,s(aq) + CH,COO. (aq) + NH4+(aq) as the precipitating reagent has the advantage of minimizing odor problems and giving denser precipitates. The four insoluble sulfides can be separated hydroxide. The sulfides of tin and antimony dissolve, forming hydroxo-complexes: into two subgroups by extracting with a solution of sodium Sas, (s) + 6 OH' (aq) ? Sn(OHA:-(aq) + 2 s-(aq) St,S, (s) + 8 OH. (aq) ? 2 Sb(OH)' (aq) + 3 S2-(aq) Since Cu" and Bi NaOH do not readily form hydroxo-complexes, CuS and Bi S, do not dissolve in solutions of complex ions is treated with HCI and thioacetamide. The solution containing the Sb(OH), and Sn(OH) The H" ions of the strong acid HCI destroy the hydroxo-complexes; the free cations then reprecipitate as the Sn(OH)''aq) + 6 H'aq) ? sn', (aq) + 6 H,O(1) SnOH)--(aq)+ 2 H' (aq) +2H2S(aq)? SnS:(s) + 6H,01) The Sb(OH). ion behaves in a very similar manner, being converted first to Sb and then to Sb S, sulfides. The reaction with Sn(OH) may be written as: The Sb S, and SnS, are then dissolved as chloro-complexes in hydrochloric acid and their presence is con- firmed by appropriate tests. The confirmatory test for tin takes advantage of the two oxidation states, +2 and +4, of the metal. Aluminum is added to reduce Sn to Sn The classic test for the presence of Sn2" in this solution is to add a solution containing mercury(I) chloride, HgCl, which would bring about another oxidation-reduction process: white

Sprint 7:32 PM @ 41% 1 a loncapa.nau.edu Reagent Matching Match the statement to the correct answer. (Some reagents can be used more than once.) The flowchart along with the report sheets in experiment 5 should be very helpful A OH, CO, and PO4 B Disodium EDTA (NazEDTA) C 0.1M sodium carbonate (Na2CO) D Lead chloride (PbCl2(s)) E 6M Hydrochloric Acid (HCI) F 1M ammonium hydroxide (NH40H) G Zinc Nitrate (Zn(NO3);2 H Silver Nitrate (AgNO3) I Silver chloride (AgCl(s)) J Anion K NH4 (aq) and OH(aq) L Colbalt Nitrate (Co(NO3)2) M Lead Nitrate (0.1M Pb(NO3)s) N Cation O Nar, NO, and NH4 P Aluminum Nitrate (Al(NO3)3) Q 6M Nitric Acid (HNO3ã R Barium Nitrate (Ba(NO3)2) S None of the above This reagent is added to fresh solutions in test B2 where there was no reaction in test B1. Salts containing these ions are often insoluble in water and form precipitates Mixing AgNO, and NaCI will produce what white precipitate In test A2 this reagent added to a fresh sample of solution where there was no reaction in test A1 In test B1 this is the first reagent added to the sample This reagent is added fresh samples in test A3 where there was no reaction in tests A1 or A2 Submt AnseTries 0/99 Threaded View Chronological View Sorting/Filtering options Export? s on what is marked as NEW Mark NEW posts no longer new reagent matching Mohamed Abu Khamiss (mya29:nau) Reply (Wed Oct 25 12:55:40 am 2017 (MST)) NEW it is hard for me to understand what exactly needed from me to answer you over did the multiple choice and the questions are more complicated than the choices it self i answered in many different forms never in my life i stay answering a question for 7 hours and it should be more clarified to the student. NEW Anonymous 2 Reply (Wed Oct 25 01:02:03 pm 2017 (MST)) Try your flow chart It's easier to look back into your lab manual. It's on page 69. Good Luck! Current discussion settings: 1. Display AW posts 2. Not new Once marked not NEW Change? Threaded View Chronological View Sorting/Filtering options Export? Preferences on what is marked as NEW Mark NEW posts no longer new