Yield of Alum Answers
Chemistry Conversion Help Please?
Q. Can someone tell me how to calculate the theoretical yield of Alum KAL (SO4)2 * 12 H2O. I am beginning with .94 grams of Aluminum and need to convert to grams of Alum but am not sure what to do?
Asked by Aye Aye Aye need help - Sun Oct 4 10:29:23 2009 - Chemistry - 1 Answers - Comments
A. molecular weight of your compound, the long thing in your question is known as aluminum, is 258.207 g. msg me bc i am confused on what you need to calculate...you didn't give enough info.
Answered by spikegirl9999 - Sun Oct 4 11:06:15 2009
Q. Can someone tell me how to calculate the theoretical yield of Alum KAL (SO4)2 * 12 H2O. I am beginning with .94 grams of Aluminum and need to convert to grams of Alum but am not sure what to do?
Asked by Aye Aye Aye need help - Sun Oct 4 10:29:23 2009 - Chemistry - 1 Answers - Comments
A. molecular weight of your compound, the long thing in your question is known as aluminum, is 258.207 g. msg me bc i am confused on what you need to calculate...you didn't give enough info.
Answered by spikegirl9999 - Sun Oct 4 11:06:15 2009
In lab we are doing synthesis of Alum. Is there a way to calculate actual yield if your substance never dried?
Q. I used 1.20g of Al, 50m L of 1.5M KOH, 20ml of 9M H2SO4. The mass of my empty beaker was 112.87g and mass of beaker and alum of 126.50g. Hope somebody can help
Asked by Ryan - Mon Oct 18 22:37:31 2010 - Chemistry - 1 Answers - Comments
A. There is no simple way to calculate your actual yield if you never dried the product. Theoretically you get 258g alum from 27g Al, so you can use simple proportion to calculate how much a 100% yield would be from the 1.2g Al you started with. Some of the alum is always going to remain in solution, so you'll not get 100% of what you expect. Guess you'll get. say 93% ...and work that out. But wouldn't it be far, far simpler just to ask somebody else in your class what they got and use a figure nearly the same? Or just confess the truth in your lab report write up?
Answered by Colin - Tue Oct 19 06:57:15 2010
Q. I used 1.20g of Al, 50m L of 1.5M KOH, 20ml of 9M H2SO4. The mass of my empty beaker was 112.87g and mass of beaker and alum of 126.50g. Hope somebody can help
Asked by Ryan - Mon Oct 18 22:37:31 2010 - Chemistry - 1 Answers - Comments
A. There is no simple way to calculate your actual yield if you never dried the product. Theoretically you get 258g alum from 27g Al, so you can use simple proportion to calculate how much a 100% yield would be from the 1.2g Al you started with. Some of the alum is always going to remain in solution, so you'll not get 100% of what you expect. Guess you'll get. say 93% ...and work that out. But wouldn't it be far, far simpler just to ask somebody else in your class what they got and use a figure nearly the same? Or just confess the truth in your lab report write up?
Answered by Colin - Tue Oct 19 06:57:15 2010
Theoretical and percentage yield?
Q. a) Assuming that aluminum foil is 100% aluminum and that is the limiting reagent, what is the theoretical yield of potassium alum if 5.00g of aluminum foil is reacted? b) If the actual yield of Alum was 50.0g, what is the percentage yield for the reaction? wow... im an idiot. hahaha heres the equation: 2Al(s) + 2KOH(aq) + 6H2O(l) --> 2KAl(OH)4(aq) 3H2(g)
Asked by Michelle - Wed Feb 2 16:54:40 2011 - Chemistry - 1 Answers - Comments
Q. a) Assuming that aluminum foil is 100% aluminum and that is the limiting reagent, what is the theoretical yield of potassium alum if 5.00g of aluminum foil is reacted? b) If the actual yield of Alum was 50.0g, what is the percentage yield for the reaction? wow... im an idiot. hahaha heres the equation: 2Al(s) + 2KOH(aq) + 6H2O(l) --> 2KAl(OH)4(aq) 3H2(g)
Asked by Michelle - Wed Feb 2 16:54:40 2011 - Chemistry - 1 Answers - Comments
How many grams of alum can be obtained from 20.0g of aluminum when the reaction proceeds with 100% yield? 80%?
Q. Need help!!! Thanks : )
Asked by Polish - Thu Oct 18 12:25:18 2007 - Mathematics - 1 Answers - Comments
A. What do you mean by alum? Aluminum sulfate? Anhydrous aluminum sulfate has a chemical formula of Al2(SO4)3. Thus it has a molecular weight of 27*2 + 3*(32 + 4*16) = 342 So 100% reaction would yield 20 * 342/54 = 126.7 g An 80% reaction would yield 20 * 342/54 * 0.8 = 101.3 g
Answered by gebobs - Thu Oct 18 12:33:22 2007
Q. Need help!!! Thanks : )
Asked by Polish - Thu Oct 18 12:25:18 2007 - Mathematics - 1 Answers - Comments
A. What do you mean by alum? Aluminum sulfate? Anhydrous aluminum sulfate has a chemical formula of Al2(SO4)3. Thus it has a molecular weight of 27*2 + 3*(32 + 4*16) = 342 So 100% reaction would yield 20 * 342/54 = 126.7 g An 80% reaction would yield 20 * 342/54 * 0.8 = 101.3 g
Answered by gebobs - Thu Oct 18 12:33:22 2007
Theoretical and Percentage Yield Help?
Q. Doing Practice Problems and Im getting this one wrong. 2Al(s) + 2KOH(aq) + 6H2O(l) --> 2KAl(OH)4(aq) 3H2(g) a) Assuming that aluminum foil is 100% aluminum and that is the limiting reagent, what is the theoretical yield of potassium alum if 5.00g of aluminum foil is reacted? b) If the actual yield of Alum was 50.0g, what is the percentage yield for the reaction?
Asked by Ac - Tue Sep 6 11:22:26 2011 - Chemistry - 1 Answers - Comments
A. from the balanced equation: 1 mol Al will produce 1 mol KAl(OH)4 Molar mass Al = 26.98g/mol 5g Al = 5/26.98 = 0.1853 mol Al This will produce 0.1853 mol KAL(OH)4 Molar mass KAl(OH)4 = 134.1094 g/mol 0.1853 mol = 0.1853*134.1094 = 24.85g theoretical yield. If the actual yield is 50g, then the % yield is approx 200% which is not possible. There is a mistake somewhere. Please check your submitted data.
Answered by Trevor H - Tue Sep 6 13:17:54 2011
Q. Doing Practice Problems and Im getting this one wrong. 2Al(s) + 2KOH(aq) + 6H2O(l) --> 2KAl(OH)4(aq) 3H2(g) a) Assuming that aluminum foil is 100% aluminum and that is the limiting reagent, what is the theoretical yield of potassium alum if 5.00g of aluminum foil is reacted? b) If the actual yield of Alum was 50.0g, what is the percentage yield for the reaction?
Asked by Ac - Tue Sep 6 11:22:26 2011 - Chemistry - 1 Answers - Comments
A. from the balanced equation: 1 mol Al will produce 1 mol KAl(OH)4 Molar mass Al = 26.98g/mol 5g Al = 5/26.98 = 0.1853 mol Al This will produce 0.1853 mol KAL(OH)4 Molar mass KAl(OH)4 = 134.1094 g/mol 0.1853 mol = 0.1853*134.1094 = 24.85g theoretical yield. If the actual yield is 50g, then the % yield is approx 200% which is not possible. There is a mistake somewhere. Please check your submitted data.
Answered by Trevor H - Tue Sep 6 13:17:54 2011
very difficult Chemistry problem?
Q. You produced 12.5 grams of alum, (KAl(SO4))2 . 12H2O) in your experiment. (a) how many moles of alum did you produce (b) What is the theoretical yield of alum if you started with 2.25g of aluminum? ( the production of alum from aluminum is a 1:1 mole ratio) (c) what is the percent yield? (d) What is percent error ?
Asked by Zatz - Tue Apr 15 13:49:54 2008 - Chemistry - 1 Answers - Comments
A. Given 12.5 g of alum part a count the atoms of each element. (2K, 2Al, 2S, 8O) + (24H, 12O) Calculate the molecular weight. Molar Mass = (2 x 39.098) + (2 x 26.9815) + (2 x 32.066) + (8 x 15.9994) + (24 x 1.00794) + (12 x 15.9994) = (78.196) + (53.963) + (64.132) + (127.9952) + (24.19056) + (191.9928) = 540.470 grams/mole However, we have 25 grams of alum... So to get the number of moles of alum... this is how we find it... 25g of alum x (1mole/540.470g) = 0.046256 moles PART B: They tell us that the production of alum from aluminum is a 1:1 mole ratio. That means that it takes one mole of Al to make one mole of the alum. They want us to find the theoretical yield of alum if you started with 2.25 g of aluminum. Since we're comparing mole… [cont.]
Answered by blueskies - Tue Apr 15 15:03:27 2008
Q. You produced 12.5 grams of alum, (KAl(SO4))2 . 12H2O) in your experiment. (a) how many moles of alum did you produce (b) What is the theoretical yield of alum if you started with 2.25g of aluminum? ( the production of alum from aluminum is a 1:1 mole ratio) (c) what is the percent yield? (d) What is percent error ?
Asked by Zatz - Tue Apr 15 13:49:54 2008 - Chemistry - 1 Answers - Comments
A. Given 12.5 g of alum part a count the atoms of each element. (2K, 2Al, 2S, 8O) + (24H, 12O) Calculate the molecular weight. Molar Mass = (2 x 39.098) + (2 x 26.9815) + (2 x 32.066) + (8 x 15.9994) + (24 x 1.00794) + (12 x 15.9994) = (78.196) + (53.963) + (64.132) + (127.9952) + (24.19056) + (191.9928) = 540.470 grams/mole However, we have 25 grams of alum... So to get the number of moles of alum... this is how we find it... 25g of alum x (1mole/540.470g) = 0.046256 moles PART B: They tell us that the production of alum from aluminum is a 1:1 mole ratio. That means that it takes one mole of Al to make one mole of the alum. They want us to find the theoretical yield of alum if you started with 2.25 g of aluminum. Since we're comparing mole… [cont.]
Answered by blueskies - Tue Apr 15 15:03:27 2008
Can someone help me with this problem in the theoretical yield?
Q. A mass of 13.02 g of (NH4)2SO4 is dissolved in water. After the solution is heated, 27.22 g of Al2(SO4)3 * 18H2O is added. Calculate the theoretical yield of the resulting alum. This is a limiting reactant problem.
Asked by daniel b - Wed Apr 8 21:58:51 2009 - Chemistry - 1 Answers - Comments
A. yeild = .0806
Answered by The Ugly Truth - Sun Apr 12 16:52:46 2009
Q. A mass of 13.02 g of (NH4)2SO4 is dissolved in water. After the solution is heated, 27.22 g of Al2(SO4)3 * 18H2O is added. Calculate the theoretical yield of the resulting alum. This is a limiting reactant problem.
Asked by daniel b - Wed Apr 8 21:58:51 2009 - Chemistry - 1 Answers - Comments
A. yeild = .0806
Answered by The Ugly Truth - Sun Apr 12 16:52:46 2009
how many grams of alum can be obtained from 20 g of aluminum when the reaction proceeds with 100 percent yield?
Q.
Asked by rjb2k6 - Fri Dec 4 09:38:13 2009 - Chemistry - 1 Answers - Comments
Q.
Asked by rjb2k6 - Fri Dec 4 09:38:13 2009 - Chemistry - 1 Answers - Comments
An alum is a "double salt" consisting of a monovalent cation, a trivalent ... a trivalent cation, and two sulf?
Q. An alum is a "double salt" consisting of a monovalent cation, a trivalent ... a trivalent cation, and two sulfate ions with twelve waters of hydration (waters of crystallization ) as part of the crystalline structure/ a) Are the twelve waters of hydration used to calculate the theoritical yield of the alum? explain. b) the 12 waters of hydration are "hydrated (strongly attracted)" to the metal ions in the crystalline alum structure. are the waters molecules more strongly hydrated to the monovalent cation or the trivalent cation? explain. c) what might you expect to happen to the alum if it were heated to a high temperature? explain.
Asked by Luko - Tue Jun 14 21:18:42 2011 - Homework Help - 1 Answers - Comments
A. Nobody is answering... Have a smile :-)
Answered by Lila - Sat Jun 18 18:28:49 2011
Q. An alum is a "double salt" consisting of a monovalent cation, a trivalent ... a trivalent cation, and two sulfate ions with twelve waters of hydration (waters of crystallization ) as part of the crystalline structure/ a) Are the twelve waters of hydration used to calculate the theoritical yield of the alum? explain. b) the 12 waters of hydration are "hydrated (strongly attracted)" to the metal ions in the crystalline alum structure. are the waters molecules more strongly hydrated to the monovalent cation or the trivalent cation? explain. c) what might you expect to happen to the alum if it were heated to a high temperature? explain.
Asked by Luko - Tue Jun 14 21:18:42 2011 - Homework Help - 1 Answers - Comments
A. Nobody is answering... Have a smile :-)
Answered by Lila - Sat Jun 18 18:28:49 2011
Calculate the theoretical/percent yield of 0.51 grams of aluminum?
Q. Steps: Calculate the number of moles of the limiting reactant that is used at the beginning of this experiment from its mass. From the stoichiometry of the reaction, determine the expected number of moles of the product, alum, to be recovered by the end of the synthesis. A review of the chemical equations reveals that there is a 1:1 stoichiometric ratio between the aluminum and alum. From the expected number of moles of the product, calculate the expected mass of product or the theoretical yield. Calculate the percent yield by dividing your actual yield of alum by the theoretical yield and multiplying by 100.
Asked by Ken - Mon Oct 19 00:24:19 2009 - Chemistry - 1 Answers - Comments
A. I do not require the steps but I need : 1) reaction 2) mass reactants
Answered by Dr.A - Mon Oct 19 09:22:23 2009
Q. Steps: Calculate the number of moles of the limiting reactant that is used at the beginning of this experiment from its mass. From the stoichiometry of the reaction, determine the expected number of moles of the product, alum, to be recovered by the end of the synthesis. A review of the chemical equations reveals that there is a 1:1 stoichiometric ratio between the aluminum and alum. From the expected number of moles of the product, calculate the expected mass of product or the theoretical yield. Calculate the percent yield by dividing your actual yield of alum by the theoretical yield and multiplying by 100.
Asked by Ken - Mon Oct 19 00:24:19 2009 - Chemistry - 1 Answers - Comments
A. I do not require the steps but I need : 1) reaction 2) mass reactants
Answered by Dr.A - Mon Oct 19 09:22:23 2009
Expected yield and % yield, college chemistry lab?
Q. 2Al+2KOH+4H2SO4+22H2O--->2KAl(SO4)2 *(dot) 12H2O+3H2 A student started with 0.0602g of aluminum foil. Assuming that aluminum is the limiting reagent, calculate the expected yield of alum. The student obtained 0.4822g of alum at the end of the experiment. What is the percent yield? Molar masses: Al=27g K=39.1g S=32g O=16g H=1g Assume that one aluminum can is completely reacted with oxygen to produce aluminum oxide, and that this energy can be completely converted to electrical energy. How long could this energy keep a 60 watt light bulb burning? Al=26.98 g/mole, O=16 g/mole, change of temp(triangle) Hf (Al2O3)=1670k J/mole, one can=14g. 1 watt=1J/sec
Asked by Hey! - Mon Mar 28 11:12:27 2011 - Chemistry - 1 Answers - Comments
A. 0.0602 g / 27 g/,ol = 0.00223 mol mole ratio to alum is 2 to 2 (1 to 1 ) so 0.00223 mol * 258.1 g/mol = 0.5756 g (expected) then %/100 = part/whole = 0.4822 g / 0.5756 g 83.78% 2 Al + 3/2 O2 ---> Al2O3 + 1670 k J 0.014 g / 26.98 g/mol = 0.0005189 mol Al 2 to 1 over to oxide ---> 0.00025945 mol of oxide then 0.00025945 mol * 1670 k J/mol = 0.4333 k J or 433.3 J now a proportion 60 J/s = 433.3 J / X s X = 7.22 s that is nearly 500 cans per hour !
Answered by Old Science Guy - Mon Mar 28 11:31:55 2011
Q. 2Al+2KOH+4H2SO4+22H2O--->2KAl(SO4)2 *(dot) 12H2O+3H2 A student started with 0.0602g of aluminum foil. Assuming that aluminum is the limiting reagent, calculate the expected yield of alum. The student obtained 0.4822g of alum at the end of the experiment. What is the percent yield? Molar masses: Al=27g K=39.1g S=32g O=16g H=1g Assume that one aluminum can is completely reacted with oxygen to produce aluminum oxide, and that this energy can be completely converted to electrical energy. How long could this energy keep a 60 watt light bulb burning? Al=26.98 g/mole, O=16 g/mole, change of temp(triangle) Hf (Al2O3)=1670k J/mole, one can=14g. 1 watt=1J/sec
Asked by Hey! - Mon Mar 28 11:12:27 2011 - Chemistry - 1 Answers - Comments
A. 0.0602 g / 27 g/,ol = 0.00223 mol mole ratio to alum is 2 to 2 (1 to 1 ) so 0.00223 mol * 258.1 g/mol = 0.5756 g (expected) then %/100 = part/whole = 0.4822 g / 0.5756 g 83.78% 2 Al + 3/2 O2 ---> Al2O3 + 1670 k J 0.014 g / 26.98 g/mol = 0.0005189 mol Al 2 to 1 over to oxide ---> 0.00025945 mol of oxide then 0.00025945 mol * 1670 k J/mol = 0.4333 k J or 433.3 J now a proportion 60 J/s = 433.3 J / X s X = 7.22 s that is nearly 500 cans per hour !
Answered by Old Science Guy - Mon Mar 28 11:31:55 2011
10pts !How many grams of alum? please explain?
Q. Assume that you used a 0.660g piece of pure aluminum in the systhesis of alum and obtained a 63.5% yield of alum. How many grams of alum did you obtain?
Asked by m6 - Fri Jan 30 01:36:37 2009 - Chemistry - 1 Answers - Comments
A. Moles Al = 0.660 g / 26.9815 g/mol= 0.0245 We will get 0.0245 x 63.5 / 100 = 0.0156 moles of KAl(SO4)2 * 12 H2O Mass = 0.0156 x 474.2 g/mol= 7.40 g
Answered by Dr.A - Fri Jan 30 05:51:52 2009
Q. Assume that you used a 0.660g piece of pure aluminum in the systhesis of alum and obtained a 63.5% yield of alum. How many grams of alum did you obtain?
Asked by m6 - Fri Jan 30 01:36:37 2009 - Chemistry - 1 Answers - Comments
A. Moles Al = 0.660 g / 26.9815 g/mol= 0.0245 We will get 0.0245 x 63.5 / 100 = 0.0156 moles of KAl(SO4)2 * 12 H2O Mass = 0.0156 x 474.2 g/mol= 7.40 g
Answered by Dr.A - Fri Jan 30 05:51:52 2009
need help with the following equation i can't seem to get it right?
Q. How many grams of potassium alum (KAl(SO4)2.12H2O) can we obtain from 0.53 g of aluminum foil, if we have available unlimited amounts of potassium hydroxide, sulfuric acid, and water? Or, in other words: what is the theoretical yield of potassium alum if we start with 0.53 g of aluminum and it is the limiting reagent?
Asked by cjfavon - Sun Sep 30 18:10:13 2007 - Chemistry - 1 Answers - Comments
A. Look at the formula - it contains one atom of Al - so one mole of Al will give one mole of potassium alum (in theory). How many moles of Al do we actually start with? 0.53g = 0.53/27 moles So that is how many moles of potassium alum we will form. How many grams is this? Simply multiply the number of moles by the relative formula mass [1x K,1x Al,2x S,8x O,24x H,12x O] and you have your answer.
Answered by Chemmunicator - Sun Sep 30 18:21:48 2007
Q. How many grams of potassium alum (KAl(SO4)2.12H2O) can we obtain from 0.53 g of aluminum foil, if we have available unlimited amounts of potassium hydroxide, sulfuric acid, and water? Or, in other words: what is the theoretical yield of potassium alum if we start with 0.53 g of aluminum and it is the limiting reagent?
Asked by cjfavon - Sun Sep 30 18:10:13 2007 - Chemistry - 1 Answers - Comments
A. Look at the formula - it contains one atom of Al - so one mole of Al will give one mole of potassium alum (in theory). How many moles of Al do we actually start with? 0.53g = 0.53/27 moles So that is how many moles of potassium alum we will form. How many grams is this? Simply multiply the number of moles by the relative formula mass [1x K,1x Al,2x S,8x O,24x H,12x O] and you have your answer.
Answered by Chemmunicator - Sun Sep 30 18:21:48 2007
Are the 12 waters of hydration used to calculate the theoretical yield of the alum? Explain.?
Q. An alum is a "double salt" consisting of a monovalent cation, a trivalent cation, and two sulfate ions with twelve waters of hydration (waters of crystallization) as part of the crystalline structure.
Asked by mmLL - Sat Nov 10 21:33:27 2007 - Chemistry - 1 Answers - Comments
A. Yes, you must use the complete formula mass of the compound to calculate the theoretical yield.
Answered by Dennis M - Sat Nov 10 21:37:16 2007
Q. An alum is a "double salt" consisting of a monovalent cation, a trivalent cation, and two sulfate ions with twelve waters of hydration (waters of crystallization) as part of the crystalline structure.
Asked by mmLL - Sat Nov 10 21:33:27 2007 - Chemistry - 1 Answers - Comments
A. Yes, you must use the complete formula mass of the compound to calculate the theoretical yield.
Answered by Dennis M - Sat Nov 10 21:37:16 2007
In the synthesis of alum lab why should you not expect a 100% yield of crystals?
Q.
Asked by Jessica - Sun Nov 13 16:42:34 2011 - Chemistry - 1 Answers - Comments
A. Because alum is hydrated and soluble, so you cant evaporate to dryness and have to crystallise it and some is left in solution.
Answered by Colin - Tue Nov 15 16:47:54 2011
Q.
Asked by Jessica - Sun Nov 13 16:42:34 2011 - Chemistry - 1 Answers - Comments
A. Because alum is hydrated and soluble, so you cant evaporate to dryness and have to crystallise it and some is left in solution.
Answered by Colin - Tue Nov 15 16:47:54 2011
how many grams of alum can be obtained from 20.0 g of aluminum if your percent yield was 65%?
Q.
Asked by lawlypawly - Sun Aug 14 23:37:03 2011 - Homework Help - 1 Answers - Comments
A. Divide 20g of Aluminum by molar mass of aluminum (27) and you get .74 mol Aluminum Since there's only one mol of aluminum per mol of alum, 20g (.74 mol) of Aluminum would make .74 mol of alum. Molar mass of alum is approximately 258g/mol. Multiply this by the mols of alum you have (.74) and you get approximately 191g of alum. However, since it was a 65% yield, you multiply 191 by .65. Answer = approximately 124.2g of Alum g = grams mol= moles g/mol = grams per mol
Answered by Michael - Mon Aug 15 00:05:59 2011
Q.
Asked by lawlypawly - Sun Aug 14 23:37:03 2011 - Homework Help - 1 Answers - Comments
A. Divide 20g of Aluminum by molar mass of aluminum (27) and you get .74 mol Aluminum Since there's only one mol of aluminum per mol of alum, 20g (.74 mol) of Aluminum would make .74 mol of alum. Molar mass of alum is approximately 258g/mol. Multiply this by the mols of alum you have (.74) and you get approximately 191g of alum. However, since it was a 65% yield, you multiply 191 by .65. Answer = approximately 124.2g of Alum g = grams mol= moles g/mol = grams per mol
Answered by Michael - Mon Aug 15 00:05:59 2011
AP Chemistry Lab Help: Synthesis of Alum?
Q. When the alum KAL(SO4)2*12H20 is being synthesized, why shouldn't you wash the crystals with pure water? Also, how does this question--- How does the solubility of alum in water change with temperature?--- affect the percent yield of alum? Any help is greatly appreciated!
Asked by ilovemypiccolo - Mon Sep 20 22:26:36 2010 - Chemistry - 1 Answers - Comments
A. why shouldn't you wash the crystals with pure water? because alum is extremely soluble in water! and you'd lose most of it! How does the solubility of alum in water change with temperature?--- it becomes much more soluble.
Answered by Colin - Tue Sep 21 07:37:34 2010
Q. When the alum KAL(SO4)2*12H20 is being synthesized, why shouldn't you wash the crystals with pure water? Also, how does this question--- How does the solubility of alum in water change with temperature?--- affect the percent yield of alum? Any help is greatly appreciated!
Asked by ilovemypiccolo - Mon Sep 20 22:26:36 2010 - Chemistry - 1 Answers - Comments
A. why shouldn't you wash the crystals with pure water? because alum is extremely soluble in water! and you'd lose most of it! How does the solubility of alum in water change with temperature?--- it becomes much more soluble.
Answered by Colin - Tue Sep 21 07:37:34 2010
Chemistry question help>>>>?
Q. I am doing a lab. I m stuck on a question. the question is : From one mole of aluminum, one mole of alum should be produced. calculate the expected yield of alum in grams based on the amount of aluminum which you started. I started with 0.94 g of aluminum. How do i solve this? just guide me to steps. Thanks
Asked by Buddy - Sat Feb 6 21:45:27 2010 - Chemistry - 1 Answers - Comments
A. first find the moles of Al, using its molar mass: 0.94 g Al @ 26.98 g/mol = 0.03484 moles of Al 0.03484 moles of Al produces 0.03484 moles of Alum now we multiply those moles of alum by the molar mass of alum if you made ammonium alum, (NH4)Al(SO4)2 12H2O, its molar mass is 453.33 g/mol (for the dodecahydrate) & 237.15 g/mol (for the anhydrous) if you made potassium alum, KAl(SO4)2 12(H2O). , its molar mass is 474.33 g/mol for the dodecahydrate & 258.21 g/mol for the anhydrate
Answered by Steve O - Sat Feb 6 23:14:11 2010
Q. I am doing a lab. I m stuck on a question. the question is : From one mole of aluminum, one mole of alum should be produced. calculate the expected yield of alum in grams based on the amount of aluminum which you started. I started with 0.94 g of aluminum. How do i solve this? just guide me to steps. Thanks
Asked by Buddy - Sat Feb 6 21:45:27 2010 - Chemistry - 1 Answers - Comments
A. first find the moles of Al, using its molar mass: 0.94 g Al @ 26.98 g/mol = 0.03484 moles of Al 0.03484 moles of Al produces 0.03484 moles of Alum now we multiply those moles of alum by the molar mass of alum if you made ammonium alum, (NH4)Al(SO4)2 12H2O, its molar mass is 453.33 g/mol (for the dodecahydrate) & 237.15 g/mol (for the anhydrous) if you made potassium alum, KAl(SO4)2 12(H2O). , its molar mass is 474.33 g/mol for the dodecahydrate & 258.21 g/mol for the anhydrate
Answered by Steve O - Sat Feb 6 23:14:11 2010
Theoretical Yield?
Q. What is the theoretical yield of potassium chromium alum (Molar Mass= 499.5)that can ve made from 2.00g of chromium (Cr)? Assume all other reagents are present in excess. If 15.0g is actually produced,what is the percent yield?(Atomic weight Cr= 52.00) Theoretical Yield=? Percent Yield=? Formula of Alum=? Pls,not only answers,clearly solution is appreciated... Thanx! Thanks for the first-2 answers, but i don't know how to get the formula of Alum.Sure, i can not search online,when i am having test...there should be a solution for it...thnx
Asked by Tommy - Mon Dec 11 09:07:59 2006 - Chemistry - 4 Answers - 1 Comments
A. Potassium chromium alum = KCr(SO4)2 * 12H2O Molar mass: 39.10 + 52.00 + (2*32.07) + (8*16.00) + (24*1.01) + (12*16.00) = 499.49 g/mol [I just want to verify the formula & yay! it's correct!] If all the other reagents are in excess, we will only have to take into consideration of how much chromium is given. In this case, we have 2.00g of Cr and that would be our limiting reagent (obviously ^_^). As we can see from the formula of the alum, it takes one mole of Cr to make one mole of the alum (yep! that would be their ratio, 1:1) Since we're comparing mole to mole, we would have to convert our given (info; 2.00g of Cr) into mole. How many mole of Cr is 2.00g of Cr? To find that, we will have to divide the given (2.00g of Cr) by 52.00 g/mol [cont.]
Answered by Cinna B - Mon Dec 11 13:26:40 2006
Q. What is the theoretical yield of potassium chromium alum (Molar Mass= 499.5)that can ve made from 2.00g of chromium (Cr)? Assume all other reagents are present in excess. If 15.0g is actually produced,what is the percent yield?(Atomic weight Cr= 52.00) Theoretical Yield=? Percent Yield=? Formula of Alum=? Pls,not only answers,clearly solution is appreciated... Thanx! Thanks for the first-2 answers, but i don't know how to get the formula of Alum.Sure, i can not search online,when i am having test...there should be a solution for it...thnx
Asked by Tommy - Mon Dec 11 09:07:59 2006 - Chemistry - 4 Answers - 1 Comments
A. Potassium chromium alum = KCr(SO4)2 * 12H2O Molar mass: 39.10 + 52.00 + (2*32.07) + (8*16.00) + (24*1.01) + (12*16.00) = 499.49 g/mol [I just want to verify the formula & yay! it's correct!] If all the other reagents are in excess, we will only have to take into consideration of how much chromium is given. In this case, we have 2.00g of Cr and that would be our limiting reagent (obviously ^_^). As we can see from the formula of the alum, it takes one mole of Cr to make one mole of the alum (yep! that would be their ratio, 1:1) Since we're comparing mole to mole, we would have to convert our given (info; 2.00g of Cr) into mole. How many mole of Cr is 2.00g of Cr? To find that, we will have to divide the given (2.00g of Cr) by 52.00 g/mol [cont.]
Answered by Cinna B - Mon Dec 11 13:26:40 2006
How many grams of dry alum can be obtained from 20.0g of aluminum when the reaction proceeds with 60.0% yield?
Q. B) How many grams of wet alum would be obtained if the reaction were to proceed with 60.0% yield but for every 4 grams of alum obtained there is one gram of excess water isolated with it?
Asked by leighlynn08 - Wed Oct 29 01:55:19 2008 - Chemistry - 1 Answers - Comments
A. 9.6g
Answered by Sgt Froggy - Wed Oct 29 02:34:59 2008
Q. B) How many grams of wet alum would be obtained if the reaction were to proceed with 60.0% yield but for every 4 grams of alum obtained there is one gram of excess water isolated with it?
Asked by leighlynn08 - Wed Oct 29 01:55:19 2008 - Chemistry - 1 Answers - Comments
A. 9.6g
Answered by Sgt Froggy - Wed Oct 29 02:34:59 2008
From Yahoo Answer Search: 'yield of alum'
Thu Jan 12 17:44:09 2012