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Supporting calculations: The net reaction is: 2Al(s) + 2KOH + 22H20 + 4H2SO4 → 2KAI(SO4)2:12H20...
how do i find theoretical yield of alum and the percent yield of alum Mass of...
how do i find theoretical yield of alum and the percent yield
of alum
Mass of aluminum foil Mass of 50 mL beaker + alum Mass of 50 mL beaker 1.019 40.99 30.35 10.59 88*C-1030 732-96c Mass of alum Melting point range of pure alum Melting point range of your alum Theoretical yield of alum Percent yield of alum Supporting calculations: The net reaction is: 2Al(s) + 2KOH + 22H2O + 4H2SO4 2KAI(SO4)2-12H2O + 3H2 Theoretical yield of alum Percent...
3. How many grams of Alum could form (theoretical yield) when 1.00 g Al (limiting reactant)...
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)
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...
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
from burde Mass of aluminum foil Mass of 50 ml beaker + alum Mass of 50...
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
Post La DUI LUI U UUDLUNS What is the biggest factor that causes vields of alum...
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...
Prelab Questions: Name 1) What is the molar mass of KAl(SO4)2 12H20 ? (the 12 waters...
Prelab Questions: Name 1) What is the molar mass of KAl(SO4)2 12H20 ? (the 12 waters are part of the molecule) 2) Calculate the mass of alum that can be produced from 14.471g of Al 3) Calculate the mass of alum that can be formed from 0.724 moles of KOH 4) Calculate the mass of alum that can be formed from 1.10 mole of H2SO4 5) a) Based on the data above which reactant is the limiting reactant? b) What...
Synthesis of Alum (Potassium Aluminum Sulfate) from Aluminum Aluminum cans can be recycled to make potassium...
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...
percent error #3 Calculations for Part 2 Ou in KAl(SO4)2 1. Experimental value for percent sulfate...
percent error #3
Calculations for Part 2 Ou in KAl(SO4)2 1. Experimental value for percent sulfate in alum BA 0.15793504) H20 12 ( 18.0a)= 216.24 KAYS0y), 12H20 MM (504)₂ = 64.327 128.00 2 = 192.32 glmor 2. Theoretical value for percent sulfate in alum are monologa dobar odbud 3. Percent error Results and Data Analysis Table 1: Determination of the Water of Hydration in Alum Crystals Mass of crucible + cover + alum 43.10509 - water Mass of crucible +...
This question has to do with an alum synthesis experiment from aluminum, KOH, and H2SO4. The...
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...
can someone help answer this please? Experiment 2 - Preparation of Alum Evaluation Question 1 Part...
can someone help answer this please?
Experiment 2 - Preparation of Alum Evaluation Question 1 Part (a) Calculate the volume of 1.0 M KOH solution required to completely dissolve 0.5589 g of aluminium. The balanced chemi follows: 2Al(s) + 2KOH(aq) + 6H2O(0) -> 2Al(OH)4 (aq) + 2K+ (aq) + 3H2(g) (Please give your answer to 2 significant figures.) mL KOH Check Answer Part (b) The 0.5589 g of aluminium metal from Part (a), reacted with excess potassium hydroxide to produce...
how do i find theoretical yield of alum and the percent yield
of alum
Mass of aluminum foil Mass of 50 mL beaker + alum Mass of 50 mL beaker 1.019 40.99 30.35 10.59 88*C-1030 732-96c Mass of alum Melting point range of pure alum Melting point range of your alum Theoretical yield of alum Percent yield of alum Supporting calculations: The net reaction is: 2Al(s) + 2KOH + 22H2O + 4H2SO4 2KAI(SO4)2-12H2O + 3H2 Theoretical yield of alum Percent...
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
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...
Prelab Questions: Name 1) What is the molar mass of KAl(SO4)2 12H20 ? (the 12 waters are part of the molecule) 2) Calculate the mass of alum that can be produced from 14.471g of Al 3) Calculate the mass of alum that can be formed from 0.724 moles of KOH 4) Calculate the mass of alum that can be formed from 1.10 mole of H2SO4 5) a) Based on the data above which reactant is the limiting reactant? b) What...
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...
percent error #3
Calculations for Part 2 Ou in KAl(SO4)2 1. Experimental value for percent sulfate in alum BA 0.15793504) H20 12 ( 18.0a)= 216.24 KAYS0y), 12H20 MM (504)₂ = 64.327 128.00 2 = 192.32 glmor 2. Theoretical value for percent sulfate in alum are monologa dobar odbud 3. Percent error Results and Data Analysis Table 1: Determination of the Water of Hydration in Alum Crystals Mass of crucible + cover + alum 43.10509 - water Mass of crucible +...
can someone help answer this please?
Experiment 2 - Preparation of Alum Evaluation Question 1 Part (a) Calculate the volume of 1.0 M KOH solution required to completely dissolve 0.5589 g of aluminium. The balanced chemi follows: 2Al(s) + 2KOH(aq) + 6H2O(0) -> 2Al(OH)4 (aq) + 2K+ (aq) + 3H2(g) (Please give your answer to 2 significant figures.) mL KOH Check Answer Part (b) The 0.5589 g of aluminium metal from Part (a), reacted with excess potassium hydroxide to produce...
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