1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Stoichiometry again (using solutions)

  1. Jul 12, 2009 #1
    1. The problem statement, all variables and given/known data

    H2SO4 + Na2CO3 g Na2SO4 + H2O + CO2

    Calculate the molarity of the H2SO4 solution if it takes 40.0 mL of H2SO4 to neutralize 46.7 mL of a 0.364 M Na2CO3 solution.


    3. The attempt at a solution

    The equation is balanced. My plan was

    grams Na2CO3 --> mols Na2CO3--> molsH2SO4---> mols/litre H2SO4(equals Molarity).


    my calculations...

    .364gramsNa2CO3 x 1molNa2CO3/106gramsNa2CO3 x 1molH2SO4/1molNa2CO3 x 1000mlH2SO4/1LH2SO4

    I know I am going wrong with the last bit but can't think of any other way to do it.

    Help!?
     
  2. jcsd
  3. Jul 12, 2009 #2

    symbolipoint

    User Avatar
    Homework Helper
    Education Advisor
    Gold Member

    At least you recognized the 1:1 mole ration between the acid and the carbonate.

    moles Na2CO3=(0.467L)(0.364M)=moles of H2SO4.

    Can you finish the rest? You used 40 mililiters of unknown concentration sulfuric acid.
     
  4. Jul 12, 2009 #3
    I don't know how, but I copied the wrong question here in this thread. I started with grams but there is no grams in the question. The question I was supposed to copy was this one...

    1. The problem statement, all variables and given/known data

    Given the following equation:

    H2SO4 + Na2CO3 ----> Na2SO4 + H2O + CO2

    Calculate the molarity of the H2SO4 solution if it takes 40.0 mL of H2SO4 to neutralize 0.364 g of Na2CO3.

    So my attempt at a solution is still the same. Sorry about that, I don't know how that happened but I should have rechecked before posting.

    Anyway, with this new question, where we are given grams, is my method a bit closer to correct or not? I am having difficulty with the end, converting to moles/litre.



    My answer was...
    The equation is balanced. My plan was

    grams Na2CO3 --> mols Na2CO3--> molsH2SO4---> mols/litre H2SO4(equals Molarity).


    my calculations...

    .364gramsNa2CO3 x 1molNa2CO3/106gramsNa2CO3 x 1molH2SO4/1molNa2CO3 x 1000mlH2SO4/1LH2SO4

    I know I am going wrong with the last bit but can't think of any other way to do it.

    Help!?
     
  5. Jul 12, 2009 #4
    Calculate the moles of H2SO4 that you have, right before the last part of your calculation. Since molarity = moles/liters, divide the moles of H2SO4 by 40.0mL in liters.
     
  6. Jul 12, 2009 #5
    perfect, thanks man. I understand fully now.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Stoichiometry again (using solutions)
  1. Solution Stoichiometry (Replies: 2)

Loading...