Question

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?

Answer 1

AlCl3 + 3(AgC2H3O2) -> Al(C2H3O2)3 + 3(AgCl)

Answer 2

Yes I was about to tell you to post that

Answer 3

First off recognize that

Molarity = moles/Liters

Answer 4

You have a volume 20ml of 0.500M acetate

Answer 5

Thus you can say

x moles/0.02L = 0.500M of acetate
solve for the moles

Answer 6

Look at the ratio you have 3 times the moles of silver acetate for everyone one mole of aluminum chloride

Answer 7

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

Answer 8

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

Answer 9

remember the conversion

Liters * 1000 = mL
mL/1000 = Liters

Answer 10

if you have questions about the basic algebra feel free to ask

Answer 11

Thanks a lot! i'll try this now

Answer 12

alright if you have problems show your work and I can help have lecture now though so yeah

Answer 13

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?

Answer 14

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

Answer 15

0.013L = 13mL

Answer 16

oh! that it! now i realize that.
Thanks !!

Answer 17

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

Answer 18

here are all the steps

Answer 19

remember you want to bring the x into the numerator and isolate it

Answer 20

you never want it in the denominator