Asked by Ka_Chin
Dissolving calcium hydroxide (∆Hsoln = -16.2 KJ/mol) in water is an exothermic process. How much lithium hydroxide (∆Hsoln = -23.6 KJ/mol) is needed to deliver the same amount of heat as 15.00 g of Ca(OH)2?
Hello everyone, thank you for taking a look at my question. I would be most appreciative if you could give me advice on how to solve this problem. I've exhausted myself trying to solve it and I am not getting anywhere. I'm not asking that you give me the answer, but can you tell me the steps I can take to solve it. Thank you!
Hello everyone, thank you for taking a look at my question. I would be most appreciative if you could give me advice on how to solve this problem. I've exhausted myself trying to solve it and I am not getting anywhere. I'm not asking that you give me the answer, but can you tell me the steps I can take to solve it. Thank you!
Answers
Answered by
DrBob222
mols Ca(OH)2 = grams/molar mass = approx 0.2 but you need a better number than that.
How much heat was released for the 15 g Ca(OH)2? That is
16.2 kJ/mol x approx 0.2 mol = approx 3 kJ.
So you need approx 3 kJ from the LiOH. How many mols do you need?
23.6 x # mols LiOH = approx 3.
Solve for # mols, then
grams LiOH = mols LiOH x molar mass LiOH.
How much heat was released for the 15 g Ca(OH)2? That is
16.2 kJ/mol x approx 0.2 mol = approx 3 kJ.
So you need approx 3 kJ from the LiOH. How many mols do you need?
23.6 x # mols LiOH = approx 3.
Solve for # mols, then
grams LiOH = mols LiOH x molar mass LiOH.
Answered by
Ka_Chin
Thank you Dr. Bob.
This is my work:
15g / 74.093 g Ca(OH)2 = 0.202
16.2 kJ/mol * 0.202 = 3.272 kJ
23.6 kJ/mol * X mol LiOH = 3.272 KJ
X mol LiOH = 0.138
0.138 * 23.95 = 3.32 g LiOH
This is my work:
15g / 74.093 g Ca(OH)2 = 0.202
16.2 kJ/mol * 0.202 = 3.272 kJ
23.6 kJ/mol * X mol LiOH = 3.272 KJ
X mol LiOH = 0.138
0.138 * 23.95 = 3.32 g LiOH
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