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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?
 one year ago
 one year 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?
 one year ago
 one year ago

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JoãoVitorMCBest ResponseYou've already chosen the best response.0
AlCl3 + 3(AgC2H3O2) > Al(C2H3O2)3 + 3(AgCl)
 one year ago

AustralopithecusBest ResponseYou've already chosen the best response.1
Yes I was about to tell you to post that
 one year ago

AustralopithecusBest ResponseYou've already chosen the best response.1
First off recognize that Molarity = moles/Liters
 one year ago

AustralopithecusBest ResponseYou've already chosen the best response.1
You have a volume 20ml of 0.500M acetate
 one year ago

AustralopithecusBest ResponseYou've already chosen the best response.1
Thus you can say x moles/0.02L = 0.500M of acetate solve for the moles
 one year ago

AustralopithecusBest 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
 one year ago

AustralopithecusBest 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
 one year ago

AustralopithecusBest 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
 one year ago

AustralopithecusBest ResponseYou've already chosen the best response.1
remember the conversion Liters * 1000 = mL mL/1000 = Liters
 one year ago

AustralopithecusBest ResponseYou've already chosen the best response.1
if you have questions about the basic algebra feel free to ask
 one year ago

JoãoVitorMCBest ResponseYou've already chosen the best response.0
Thanks a lot! i'll try this now
 one year ago

AustralopithecusBest 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
 one year ago

JoãoVitorMCBest 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?
 one year ago

AustralopithecusBest 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
 one year ago

AustralopithecusBest ResponseYou've already chosen the best response.1
0.013L = 13mL
 one year ago

JoãoVitorMCBest ResponseYou've already chosen the best response.0
oh! that it! now i realize that. Thanks !!
 one year ago

AustralopithecusBest 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
 one year ago

AustralopithecusBest ResponseYou've already chosen the best response.1
here are all the steps
 one year ago

AustralopithecusBest ResponseYou've already chosen the best response.1
remember you want to bring the x into the numerator and isolate it
 one year ago

AustralopithecusBest ResponseYou've already chosen the best response.1
you never want it in the denominator
 one year ago
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