A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

anonymous

  • one year ago

What volume of 1.2M H2SO4 is required to react exactly with 3.5g of NaOH (MW=40.0g/mol) according to the following reaction? 2NaOH+H2SO4---> NaSO4+2H2O

  • This Question is Closed
  1. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I am guessing it's another M1V1=M2V2 equation.

  2. aaronq
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    nope, this is a stoichiometry problem not a dilution (you're reacting a base and an acid) have you done these before?

  3. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    the better questions is did I understand them to begin with and have I asked my teacher? question one no and question two yes (wasn't much help since he quoted the book which I didn't understand in the first place)

  4. aaronq
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    okay, do you want me tell you how, or you're okay with it?

  5. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    please explain away.

  6. aaronq
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    lol alright. I have this written out from before. General Scheme: \(\sf \large 1.\)First write and balance an equation for the process described. So you have the chemical equation balanced (this is crucial because you need the stoichiometric coefficients). \(\sf \large 2.\)Next, use the stoichiometric coefficients to find moles produced. \(\sf \large 3.\) Set up a ratio using the species of interest, like so: e.g. for a general reaction: \(\sf \large \color{red}{a}A + \color{blue}{b}B \rightleftharpoons \color{green}{c}C\) where upper case are the species (A,B,C), and lower case (a,b,c) are the coefficients , \(\sf \dfrac{n_A}{\color{red}{a}}=\dfrac{n_B}{\color{blue}{b}}=\dfrac{n_C}{\color{green}{c}}\) From here you can isolate what you need. For example: if you have 2 moles of B, how many moles of C can you produce? solve algebraically: \(\sf\dfrac{2}{\color{blue}{b}}=\dfrac{n_C}{\color{green}{c}}\rightarrow n_C=\dfrac{2*\color{green}{c}}{\color{blue}{b}}\) \(\sf \large 4.\) Solve ---------------- you need to work with moles in stoichiometric problems, so convert the mass 3.5 g to moles of NaOH. \(\sf moles=\dfrac{mass}{Molar~mass}\) -------------- once you found moles of sulfuric acid, convert to volume using: \(\sf molarity=\dfrac{moles}{Liters }\)

  7. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Sorry I asked.

  8. aaronq
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    lol it's much simpler than it looks. finding moles of NaOH moles of NaOH \(=\dfrac{3.5~g}{39.997 g/mol} = 0.08750~moles \) Using stoichiometric coefficients in ratio with moles \(\sf \dfrac{0.08750~moles }{1}=\dfrac{moles~ of~ H_2SO_4}{2}\) \(\sf moles~ of~ H_2SO_4=2*\dfrac{0.08750~moles }{1}=0.1750~ moles\) Finding volume of sulfuric acid \(\sf L=\dfrac{0.1750~ moles}{1.2~M}=0.1458~L=145.8~mL\approx 146 mL\)

  9. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Were did you pull the 2 from?

  10. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Where*

  11. aaronq
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    the 2 is from the coefficients,...oh crap, i messed it. it should be \(\sf \dfrac{0.08750~moles }{2}=\dfrac{moles~ of~ H_2SO_4}{1}\)

  12. aaronq
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    so then \(\sf moles~ of~ H_2SO_4=1*\dfrac{0.08750~moles }{2}=0.04375~ moles\)

  13. aaronq
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    and \(\sf L=\dfrac{0.04375~ moles}{1.2~M}=0.03645~L=36.5~mL\)

  14. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    This is going to be just a rhetorical question but really what kind of stupid insanity is behind all that. It's like getting a math problem you won't even use in real life.

  15. aaronq
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    this is a very general problem, it'll come across in real life if you do any kind of work with wet chemistry

  16. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    None because I am a friggen Mechanic.

  17. aaronq
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    yeah, then i guess it wont.

  18. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Ok looks like I got another problem that looks like this one. Zinc dissolves in hydro chloric acid to yield to hydrogen gas: Zn(s)+2HCl(aq)----> ZnCl2(aq)+H2(g). When a 12.7g sample of zinc dissolves in 5.00*10^2ml of 1.450M HCL. What is the concentration of chloride ion in the final solution?

  19. aaronq
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    sort of, but you just have find the molarity of ZnCl2 in the solution, then multiply that by 2 (because there are two Cl^- ions for every Zn^(2+) ion \(\sf ZnCl_2\rightarrow Zn^{2+}+2Cl^-\) moles of MgCl2= 12.7g/95.211 g/mol=0.1333 moles \(molarity~of~MgCl_2=\dfrac{0.1333~moles}{0.5~L}=0.2667~M\) \([Cl^-]=2*[MgCl_2]=2*0.2667~=0.534~M\)

  20. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    doesn't equal any of the possible answers. Possible Answers .674 mole .776 mole 1.06 moles 1.45 moles

  21. aaronq
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    sorry, i keep reading the questions, wrong. you used HCl not ZnCl2, so the concentration of \(Cl^-\) is same as that of HCl, because they are 1:1 in number of atoms.

  22. aaronq
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    \(HCl\rightarrow H^++Cl^-\) so [Cl^-] =1.450 M

  23. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I friggen hate chemistry. Thank you again.

  24. aaronq
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    it gets much more interesting once you get past this basic general chemistry stuff. no problem

  25. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @aarong I just retook the quiz I was taking and looked for that 1.45M answer and it wasn't there. The remaining answers that where there that was in the previous questions where .674M, .776M, and 1.06M. The answer turned out to be .674M. Care to explain?

  26. aaronq
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Are you sure it was the exact same question? the answer \(is\) 1.45 M

  27. aaronq
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    The volume didn't change, so the molarity wouldn't have changed.

  28. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Dude i checked three times before answering.

  29. aaronq
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    that's really strange because that is the answer, hold on let me try to google the question

  30. aaronq
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    I can't find anything. sorry dude. I know the concentration of chloride ion would be the same because both HCl or ZnCl2 are very soluble in water, so they wouldn't precipitate out of solution thus remain unchanged throughout the reaction.

  31. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    It is what it is. Frankly I think these quizzes are messed up beyond belief.

  32. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.