Stoichiometry again (using solutions)

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The discussion focuses on calculating the molarity of an H2SO4 solution needed to neutralize a Na2CO3 solution. The balanced chemical equation is provided, and the user initially misinterprets the problem by using grams instead of moles. After clarification, the correct approach involves converting grams of Na2CO3 to moles, then using the stoichiometric ratio to find moles of H2SO4. The final step requires dividing the moles of H2SO4 by the volume of the solution in liters to find the molarity. The user expresses gratitude for the clarification and confirms their understanding of the calculation process.
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Homework Statement



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.


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!?
 
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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.
 
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. Homework Statement

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!?
 
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.
 
perfect, thanks man. I understand fully now.
 

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