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anonymous
 4 years ago
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?
anonymous
 4 years ago
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?

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0AlCl3 + 3(AgC2H3O2) > Al(C2H3O2)3 + 3(AgCl)

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

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

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

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

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

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

Australopithecus
 4 years ago
Best ResponseYou've already chosen the best response.1This 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

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

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Thanks a lot! i'll try this now

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i 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?

Australopithecus
 4 years ago
Best ResponseYou've already chosen the best response.10.500M = 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh! that it! now i realize that. Thanks !!

Australopithecus
 4 years ago
Best ResponseYou've already chosen the best response.1I 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

Australopithecus
 4 years ago
Best ResponseYou've already chosen the best response.1here are all the steps

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

Australopithecus
 4 years ago
Best ResponseYou've already chosen the best response.1you never want it in the denominator
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