how do i find theoretical yield of alum and the percent yield of alum Mass of...
3. How many grams of Alum could form (theoretical yield) when 1.00 g Al (limiting reactant) is used according to the balanced chemical equation given below? 2Al(s) + 2KOH(aq) + 4H2SO4 + 22H2O(1) ► 2KAI(SO4)2:12H2O(s) + 3H2(g)
from burde Mass of aluminum foil Mass of 50 ml beaker + alum Mass of 50 mL beaker 0.183 96.1 95 20g 10.96 Mass of alum Melting point range of pure alum Melting point range of your alum Theoretical yield of alum Percent yield of alum 91-4°C 74.0-9°C Supporting calculations: The net reaction is: 2Al(s) + 2KOH + 22H2O + 4H.SO K AKSO )2 12H20 + 3H2 Theoretical yield of alum 0.98 Ilmol 1 30.565 Percent yield of alum
2Al(s)+2KOH(aq)+4H2SO4+22H2O(l)→2KAl(SO4)2⦁12H2O(s)+3H2(g) What is the limiting reactant, theoretical and percent yield of this equation? 1.03 grams of Al foil 35.10 grams of KOH 15 grams of Sulfuric acid 20 grams of water the mass of the product is 24.88 grams
i Supporting calculations: The net reaction is: 2Al(s) + 2KOH + 22H20 + 4H2SO4 → 2KAI(SO4)2:12H20 + 3H2 Theoretical yield of alum 1.01g| I mol All2 mol KAl(SO4)2 #12 H2O1 26.989 Al 2 mol Al 1 mol KAI(SO4)2:12H20
Post La DUI LUI U UUDLUNS What is the biggest factor that causes vields of alum in this experiment to be low? using too much Hasoy whold cause the yield to be in 2. A student reacted 1.01 g of Al with 25.0 mL of 2.00 M KOH according to the following reaction: 2Al(s) + 2KOH(aq) +6H2O(0) ► 2KAI(OH)4 + 3H2(g) a. How many moles of KOH were required to react with all of the aluminum? b. How many moles...
theoretical value for percent sulfate in alum 2. Theoretical value for percent sulfate in alum Results and Data Analysis Table 1: Determination of the water of Hydration in Alum Crystals Mass of crucible + cover + alum 43.10509_-l water Mass of crucible + cover 4.09959 Mass of alum 1.00559_ Mass of crucible + cover + alum after heating 42.65019 Mass of anhydrous alum 0.55069 Moles of anhydrous alum 0.55063 Alun 0.002132 mol nem 3 351 21) Mass of water of...
Synthesis of Alum (Potassium Aluminum Sulfate) from Aluminum Aluminum cans can be recycled to make potassium aluminum sulfate dodecahydrate, KAK(SO)12H20, which is abbreviated to "alum". The 12 waters are part of the overall formula. Alum can be used to make dyes, printing fabrics, making paper, adhesives, pickling of food and vets use it to induce a dog to throw up. x 1 rxn 2 run 3 The following are the four reactions that lead to alum: Al(s) + 2 KOH...
How do I calculate the theoretical amount of maleic anhydride, theoretical yield, and percent yield? Amount of 3-sulfolene used = 2.29g and 0.0194mol Amount of maleic anhydride used = 1.22g and 0.0124mol Mass of product isolated = 0.94g Experimental Melting Point = 87-89.5 degrees C 05 ( 137-140°C / Summary of Data Amount of 3-sulfolene used (in both grams and moles) Amount of maleic anhydride used in both grams and moles) Theoretical amount of 1,3-butadiene generated (in grams and moles)...
How do I calculate the theoretical amount of maleic anhydride, theoretical yield, and percent yield? Amount of 3-sulfolene used = 2.29g and 0.0194mol Amount of maleic anhydride used = 1.22g and 0.0124mol Mass of product isolated = 0.94g Experimental Melting Point = 87-89.5 degrees C Summary of Data Amount of 3-sulfolene used (in both grams and moles) Amount of maleic anhydride used in both grams and moles) Theoretical amount of 1,3-butadiene generated (in grams and moles) Mass of product isolated...
Перо Name _Date Section 1. mass of aluminum 16.2058 2. mass of alum possible (theoretical yield) 3. mass of alum produced (actual yield) 4. percent yield Show calculations by number: 10:52 Exit If you removed the solvent from 50.07 mL of a 1.403 M copper(II) sulfate solution, how many grams of copper(II) sulfate would you have? Question 2 3 pts The solubility of copper(II) bromide in water is 55.7 g/100 mL of water at 20 C. If you have a...