anonymous
  • anonymous
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
Chemistry
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
I am guessing it's another M1V1=M2V2 equation.
aaronq
  • aaronq
nope, this is a stoichiometry problem not a dilution (you're reacting a base and an acid) have you done these before?
anonymous
  • anonymous
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)

Looking for something else?

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

More answers

aaronq
  • aaronq
okay, do you want me tell you how, or you're okay with it?
anonymous
  • anonymous
please explain away.
aaronq
  • aaronq
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: \(\sf molarity=\dfrac{moles}{Liters }\)
anonymous
  • anonymous
Sorry I asked.
aaronq
  • aaronq
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 \(\sf L=\dfrac{0.1750~ moles}{1.2~M}=0.1458~L=145.8~mL\approx 146 mL\)
anonymous
  • anonymous
Were did you pull the 2 from?
anonymous
  • anonymous
Where*
aaronq
  • aaronq
the 2 is from the coefficients,...oh crap, i messed it. it should be \(\sf \dfrac{0.08750~moles }{2}=\dfrac{moles~ of~ H_2SO_4}{1}\)
aaronq
  • aaronq
so then \(\sf moles~ of~ H_2SO_4=1*\dfrac{0.08750~moles }{2}=0.04375~ moles\)
aaronq
  • aaronq
and \(\sf L=\dfrac{0.04375~ moles}{1.2~M}=0.03645~L=36.5~mL\)
anonymous
  • anonymous
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.
aaronq
  • aaronq
this is a very general problem, it'll come across in real life if you do any kind of work with wet chemistry
anonymous
  • anonymous
None because I am a friggen Mechanic.
aaronq
  • aaronq
yeah, then i guess it wont.
anonymous
  • anonymous
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?
aaronq
  • aaronq
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 \(molarity~of~MgCl_2=\dfrac{0.1333~moles}{0.5~L}=0.2667~M\) \([Cl^-]=2*[MgCl_2]=2*0.2667~=0.534~M\)
anonymous
  • anonymous
doesn't equal any of the possible answers. Possible Answers .674 mole .776 mole 1.06 moles 1.45 moles
aaronq
  • aaronq
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.
aaronq
  • aaronq
\(HCl\rightarrow H^++Cl^-\) so [Cl^-] =1.450 M
anonymous
  • anonymous
I friggen hate chemistry. Thank you again.
aaronq
  • aaronq
it gets much more interesting once you get past this basic general chemistry stuff. no problem
anonymous
  • anonymous
@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?
aaronq
  • aaronq
Are you sure it was the exact same question? the answer \(is\) 1.45 M
aaronq
  • aaronq
The volume didn't change, so the molarity wouldn't have changed.
anonymous
  • anonymous
Dude i checked three times before answering.
aaronq
  • aaronq
that's really strange because that is the answer, hold on let me try to google the question
aaronq
  • aaronq
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.
anonymous
  • anonymous
It is what it is. Frankly I think these quizzes are messed up beyond belief.

Looking for something else?

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