CHEM 1212 Chapter : Exp2 Synthesis Of Potassium Alum

12 views5 pages

azuremammoth471

15 Mar 2019

School

Department

Course

Professor

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

Document Summary

Alums are a class of double salts with the formula of mal(so4)2(cid:149)12 h2o. 2 anions: it is a hydrated salt, e. g. it has water molecules associated with the salt. These water molecules are included in the molar mass or molecular weight. Step 2: neutralizing the oh (aq) + 2 h2so4(aq) . Al+3 (aq) + k+ (aq) + 2 so4. To get the overall equation, you must multiply the. This will cause the intermediate species to cancel completely. 2 al(s) + 2 koh(aq) + 6 h2o( ) (aq) + 3 h2(g) + 2 k+ 2 al+3 (aq) + k+ (aq) + 2 so4. 2 al(s) + 2 koh(aq) +4 h2so4(aq) + 22 h2o( ) . Theoretical yield: in theory, the amount of product that could be produced if the reaction were. 2 mol al: find grams of potassium alum g k alum = mol alum x mwalum, or one step calculation: g k alum = gal x 1.

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

1. Potassium aluminum sulfate dodecahydrate, KAl(SO4)2Â·12H2O, is a member of the alum salts. It has a molar mass of 474 g/mol and beautiful solid crystals can be grown from aqeuos solution according to this chemical equation: K+(aq) + Al3+(aq) + 2 SO42-(aq) + 12 H2O(l) â KAl(SO4)2Â·12H2O(s) The next three questions are based on a synthesis of the alum salt product performed by Kemmi Major. To begin, Kemmi weighs 0.50 g of aluminum metal. She dissolves the aluminum with strong base to obtain Al3+ cations and adds an excess of the other reactant ions (K+ and SO42-) to the solution. She cools the solution in an ice bath until crystals of alum salt form. What is the theoretical yield of alum product in grams if the reaction works perfectly? (Hint: convert the limiting reactant aluminum to moles of Al. Assume each Al atom gives one Al3+ cation. Use the balanced chemical equation to predict the moles of alum product, and convert this to grams using the molar mass given.) Round the answer to 2 significant digits, with units of g.

2.Kemmi dries her alum crystals and weighs them. Her experimental mass is 6.7 g of alum product from her experiment. What is Kemmiâs percent yield? (Remember, the answer to the previous question is her theoretical yield.) Report 2 significant digits with units of %

a. Potassium aluminum sulfate dodecahydrate, KAl(SO₄)₂·12H₂O, is a member of the alum salts. It has a molar mass of 474 g/mol and beautiful solid crystals can be grown from an aqueous solution according to this chemical equation:

Kemmi weighs 0.50 g of aluminum metal. She dissolves the aluminum with a strong base to obtain Al³⁺ cations and adds an excess of the other reactant ions (K⁺ and SO₄^2-) to the solution. She cools the solution in an ice bath until crystals of alum salt form. What is the theoretical yield of alum product in grams if the reaction works perfectly? (Hint: convert the limiting reactant aluminum to moles of Al. Assume each Al atom gives one Al³⁺ cation. Use the balanced chemical equation to predict the moles of alum product, and convert this to grams using the molar mass given.) Round the answer to 2 significant digits, with units of g.

b. Kemmi dries her alum crystals and weighs them. Her experimental mass is 6.7 g of alum product from her experiment. What is Kemmi’s percent yield? (Remember, the answer to the previous question is her theoretical yield.) Report 2 significant digits with units of %

mauvecaribou808

CHEM 1212 Chapter : Exp2 Synthesis Of Potassium Alum

Document Summary

Get access

Related textbook solutions

Chemistry: Structure and Properties

Basic Chemistry

Principles of Chemistry Molecular Approach

Principles of Chemistry Molecular Approach

Chemistry: Structure and Properties

Chemistry: A Molecular Approach

Chemistry: A Molecular Approach

Principles of Chemistry: A Molecular Approach

Chemistry: The Central Science

Related Documents

CHEM 1212 Chapter : B64fa6ca32d2c84c56684344d9722aeb

CHEM 1212 : Chem Lab Midterm

CHEM 1212 : CHEM 1212 Midterm

Related Questions