Stoichiometry again (using solutions)

In summary: I did wrong was that I didn't divide the moles by 0.04L. Thanks again for your help and time!In summary, the molarity of the H2SO4 solution can be calculated by first converting the given grams of Na2CO3 to moles, then using the balanced equation to determine the moles of H2SO4 needed to neutralize it. Finally, divide the moles of H2SO4 by the volume of H2SO4 used in liters to find the molarity.
  • #1
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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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  • #2
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.
 
  • #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. 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!?
 
  • #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.
 
  • #5
perfect, thanks man. I understand fully now.
 

Related to Stoichiometry again (using solutions)

1. What is stoichiometry?

Stoichiometry is a branch of chemistry that deals with the quantitative relationships between the reactants and products in a chemical reaction. It involves using the mole ratio of substances to determine the amount of product produced or reactant needed.

2. How is stoichiometry used in solutions?

In solutions, stoichiometry is used to determine the amount of reactants or products involved in a chemical reaction. This is done by using the concentration of the solution and the volume of the solution to calculate the amount of substance present.

3. What is a molarity and how is it calculated?

Molarity is a unit of concentration used in stoichiometry. It is calculated by dividing the moles of solute by the volume of the solution in liters. The formula for molarity is M = moles of solute / volume of solution (in L).

4. How do you use stoichiometry to find the concentration of a solution?

To find the concentration of a solution using stoichiometry, you will need the volume and concentration of another solution that reacts with it. By using the mole ratio between the two solutions, you can calculate the concentration of the unknown solution.

5. Can stoichiometry be used to predict the outcome of a reaction?

Yes, stoichiometry can be used to predict the outcome of a reaction by calculating the amount of product that will be produced based on the given amounts of reactants. However, it is important to note that other factors such as temperature and pressure can also affect the outcome of a reaction.

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