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

- anonymous

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- anonymous

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

- aaronq

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

- 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)

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## More answers

- aaronq

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

- anonymous

please explain away.

- 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

Sorry I asked.

- 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

Were did you pull the 2 from?

- anonymous

Where*

- 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

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

- aaronq

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

- 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

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

- anonymous

None because I am a friggen Mechanic.

- aaronq

yeah, then i guess it wont.

- 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

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

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

- 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

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

- anonymous

I friggen hate chemistry. Thank you again.

- aaronq

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

- 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

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

- aaronq

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

- anonymous

Dude i checked three times before answering.

- aaronq

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

- 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

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

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