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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
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  1. JoãoVitorMC Group Title
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    AlCl3 + 3(AgC2H3O2) -> Al(C2H3O2)3 + 3(AgCl)

    • 2 years ago
  2. Australopithecus Group Title
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    Yes I was about to tell you to post that

    • 2 years ago
  3. Australopithecus Group Title
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    First off recognize that Molarity = moles/Liters

    • 2 years ago
  4. Australopithecus Group Title
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    You have a volume 20ml of 0.500M acetate

    • 2 years ago
  5. Australopithecus Group Title
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    Thus you can say x moles/0.02L = 0.500M of acetate solve for the moles

    • 2 years ago
  6. Australopithecus Group Title
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    Look at the ratio you have 3 times the moles of silver acetate for everyone one mole of aluminum chloride

    • 2 years ago
  7. Australopithecus Group Title
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    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
  8. Australopithecus Group Title
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    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
  9. Australopithecus Group Title
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    remember the conversion Liters * 1000 = mL mL/1000 = Liters

    • 2 years ago
  10. Australopithecus Group Title
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    if you have questions about the basic algebra feel free to ask

    • 2 years ago
  11. JoãoVitorMC Group Title
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    Thanks a lot! i'll try this now

    • 2 years ago
  12. Australopithecus Group Title
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    alright if you have problems show your work and I can help have lecture now though so yeah

    • 2 years ago
  13. JoãoVitorMC Group Title
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    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
  14. Australopithecus Group Title
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    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
  15. Australopithecus Group Title
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    0.013L = 13mL

    • 2 years ago
  16. JoãoVitorMC Group Title
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    oh! that it! now i realize that. Thanks !!

    • 2 years ago
  17. Australopithecus Group Title
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    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
  18. Australopithecus Group Title
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    here are all the steps

    • 2 years ago
  19. Australopithecus Group Title
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    remember you want to bring the x into the numerator and isolate it

    • 2 years ago
  20. Australopithecus Group Title
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    you never want it in the denominator

    • 2 years ago
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