mole ratio Al : ALUM from the balanced overall reaction: 2 Al(s) + 2 KOH(aq) + 22 H2O(l) + 4 H2SO4 → 2 [KAl(SO4)2 • 12 H2O(s)] + 3 H2(g)
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mole ratio Al : ALUM from the balanced overall reaction: 2 Al(s) + 2 KOH(aq) +...
Show the calculation of (1) the expected grams of alum (KAl(SO4)2 •12 H2O) formed from the reaction of 1.3 grams of aluminum metal and then show the calculation of (2) the % yield if the actual yield of alum is 19.32 grams. Overall reaction is shown below: 2 Al + 2 KOH + 10 H2O + 4 H2SO4 → 2 KAl(SO4)2 • 12 H2O + 3 H2(g)↑
Balance the following equation: Al(s) + KOH(aq) + H2SO4(aq) + H2O(l) ---> KAl(SO4)2 * 12 H2O(s) + H2(g)
During a "Synthesis of Alum" experiment, with 24.9mL of 1.4M KOH. After cooling the mixt Given the following balanced equation: OF Alum" experiment, a student uses 0.500g of aluminum foil and disson NOR. After cooling the mixture, the students adds 11.9mL of 9.OM H2SO4. 2 Al(s) + 2 KOH(aq) + 4 H,SO(aq) + 22 H200) — 3 H2(g) + 2 KAl(SO4)2-12 H20 a. Calculate the theoretical yield of alum. Show the process of finding the limiting reactant. [Molar Mass (Alum)...
This question has to do with an alum synthesis experiment from aluminum, KOH, and H2SO4. The overall reaction is given in equation (1-1): (1-1) 2Al(s) + 2KOH(aq) + 22H2O(l) + 4H2SO4(aq) ---> 2KAl(SO4)2.12H2O(s) + 3H2(g) Note: we used 0.045 moles of H2SO4 and 0.0175 moles of KOH in the experiment. Question 1: Given that KAl(OH)4 was limiting, how many moles of H2SO4 were actually used to form the maximum possible amount of KAl(SO4)2 (Hint: Moles of Al used in this...
2 Al (s) + 2 KOH (aq) + 4 H2SO4 + 22 H2O → 2 KAl(SO4)2 x 12H2O + 3H2 What is the net ionic equation?
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...
If a student needed to make 25.678 g alum and KOH and H2SO4 are in excess, how much Al should the students start with knowing that they would only get an 80% yield of alum? *There was an answer to this on Chegg, but I wasn’t too sure bc they used Al2(SO4)3 as the alum chemical formula, but I thought it was KAl (SO4)2 • 12H2O)
Which reaction is an example of an acid-base reaction? O FeCl3 (aq) + 3 KOH(aq) - Fe(OH)3 (s) + 3 KCl(aq) O 2 Hg(1) + O2(g) -2 HgO(s) O 6 HCl(aq) + 2 Al(s) -> 2 AlCl3 (aq) + 3 H2 (g) O H2SO4 (aq) + Ca(OH)2 (aq) - CaSO4 (aq) + 2 H2O(l) H2CO3 (aq) - H2O(l) + CO2 (g)
What is the net ionic equation adding KOH drop by drop and what
is the equation when you add too much?
calculate the theoretical yield for the production of alum using the following equation. 2 Al (s) + 2 KOH 22 H2O+4 H2So4 2 KAI(SO)212 H2O (s) +3 H2 (g) In Part 2 of this experiment you will qualitatively analyze the composition of alum, KAI(SO4)2-12 H2O (s), by analyzing for AB+, s042-, K+, and H2O. o When 0.1 M BaCl2...
Calculate the mass of aluminum required to fully react all of the H2SO4 in the following reaction. 2 Al(s) + 2 KOH(aq) + 22 H2O(1) + 4 H2SO4(aq) → 2 KAI(SO4)2 · 12 H2O(s) + 3 H2(g) Type your numerical answer without units.