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JoãoVitorMC
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Consider the reaction of aluminum chloride with silver acetate. How ml of 0.250M aluminum chloride are required to react completely with 20ml of 0.500M silver acetate?
 2 years ago
 2 years ago
JoãoVitorMC Group Title
Consider the reaction of aluminum chloride with silver acetate. How ml of 0.250M aluminum chloride are required to react completely with 20ml of 0.500M silver acetate?
 2 years ago
 2 years ago

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JoãoVitorMC Group TitleBest ResponseYou've already chosen the best response.0
AlCl3 + 3(AgC2H3O2) > Al(C2H3O2)3 + 3(AgCl)
 2 years ago

Australopithecus Group TitleBest ResponseYou've already chosen the best response.1
Yes I was about to tell you to post that
 2 years ago

Australopithecus Group TitleBest ResponseYou've already chosen the best response.1
First off recognize that Molarity = moles/Liters
 2 years ago

Australopithecus Group TitleBest ResponseYou've already chosen the best response.1
You have a volume 20ml of 0.500M acetate
 2 years ago

Australopithecus Group TitleBest ResponseYou've already chosen the best response.1
Thus you can say x moles/0.02L = 0.500M of acetate solve for the moles
 2 years ago

Australopithecus Group TitleBest ResponseYou've already chosen the best response.1
Look at the ratio you have 3 times the moles of silver acetate for everyone one mole of aluminum chloride
 2 years ago

Australopithecus Group TitleBest ResponseYou've already chosen the best response.1
Since the division cancels multiplication you simply divide the number of moles of silver acetate by 3 to solve for the moles of aluminum chloride
 2 years ago

Australopithecus Group TitleBest ResponseYou've already chosen the best response.1
This will give you the moles, since you have the concentration of aluminum chloride you can just sub them into the equation to solve for the liters Molarity = moles/liters
 2 years ago

Australopithecus Group TitleBest ResponseYou've already chosen the best response.1
remember the conversion Liters * 1000 = mL mL/1000 = Liters
 2 years ago

Australopithecus Group TitleBest ResponseYou've already chosen the best response.1
if you have questions about the basic algebra feel free to ask
 2 years ago

JoãoVitorMC Group TitleBest ResponseYou've already chosen the best response.0
Thanks a lot! i'll try this now
 2 years ago

Australopithecus Group TitleBest ResponseYou've already chosen the best response.1
alright if you have problems show your work and I can help have lecture now though so yeah
 2 years ago

JoãoVitorMC Group TitleBest ResponseYou've already chosen the best response.0
i did the this: 1 mole of AlCl3 > 3 moles of AgC2H3O2 x > 0.01 so x= 0.0033. 0.5 = 0.003/volume , volume = 0.0066, is it right?
 2 years ago

Australopithecus Group TitleBest ResponseYou've already chosen the best response.1
0.500M = x/0.020L x = 0.500M*0.020L x=0.01moles 0.01moles/3 = 0.0033mol of Ag acetate thus 0.250M = 0.0033mol/x 0.0033mol/0.250 = x x = 0.013L
 2 years ago

Australopithecus Group TitleBest ResponseYou've already chosen the best response.1
0.013L = 13mL
 2 years ago

JoãoVitorMC Group TitleBest ResponseYou've already chosen the best response.0
oh! that it! now i realize that. Thanks !!
 2 years ago

Australopithecus Group TitleBest ResponseYou've already chosen the best response.1
I think you made a mistake with your algebra remember Molarity = Moles/Volume thus you have 0.250M = 0.0033mol/x = 0.250M*x = 0.0033mol*x/x = 0.250M*x = 0.0033mol = 0.250M*x/0.250M = 0.0033mol/0.250M = x = 0.0033mol/0.250M
 2 years ago

Australopithecus Group TitleBest ResponseYou've already chosen the best response.1
here are all the steps
 2 years ago

Australopithecus Group TitleBest ResponseYou've already chosen the best response.1
remember you want to bring the x into the numerator and isolate it
 2 years ago

Australopithecus Group TitleBest ResponseYou've already chosen the best response.1
you never want it in the denominator
 2 years ago
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