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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?

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AlCl3 + 3(AgC2H3O2) -> Al(C2H3O2)3 + 3(AgCl)
Yes I was about to tell you to post that
First off recognize that Molarity = moles/Liters

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You have a volume 20ml of 0.500M acetate
Thus you can say x moles/0.02L = 0.500M of acetate solve for the moles
Look at the ratio you have 3 times the moles of silver acetate for everyone one mole of aluminum chloride
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
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
remember the conversion Liters * 1000 = mL mL/1000 = Liters
if you have questions about the basic algebra feel free to ask
Thanks a lot! i'll try this now
alright if you have problems show your work and I can help have lecture now though so yeah
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?
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
0.013L = 13mL
oh! that it! now i realize that. Thanks !!
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
here are all the steps
remember you want to bring the x into the numerator and isolate it
you never want it in the denominator

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