What is the amount of Sodium Carbonate in a sample after combustion?

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Discussion Overview

The discussion revolves around a chemistry problem involving a mixture of Sodium bicarbonate (NaHCO3) and Sodium carbonate (Na2CO3). Participants are trying to determine the initial amount of Sodium carbonate in a 5-gram sample after combustion at 300 Celsius, which results in a weight of 3.71 grams post-reaction. The focus is on the stoichiometric relationships and equations necessary to solve for the unknowns involved.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant outlines the reaction process and the resulting weight after combustion, indicating the need to calculate the initial amount of Sodium carbonate.
  • Another participant emphasizes the need for the original poster to show their work to receive assistance, adhering to forum guidelines.
  • A participant suggests that there are four unknowns in the problem and that four equations are necessary to solve for them, proposing a method to express all variables in terms of one unknown.
  • Further clarification is requested regarding the relationship between the mass of bicarbonate and the mass of carbonate produced, with a suggestion to use molar masses to establish this relationship.
  • Another participant reiterates the challenge of finding the link between the masses of bicarbonate and carbonate, indicating a need for additional equations based on stoichiometry.

Areas of Agreement / Disagreement

Participants generally agree on the need for additional equations to solve the problem, but there is no consensus on how to establish the necessary relationships between the variables involved.

Contextual Notes

Participants express uncertainty regarding the stoichiometric relationships and the equations needed to solve for the unknowns, indicating that assumptions about the relationships between the masses are not fully resolved.

lo2
Ok we have a mix of Sodium bicarbonate (NaHCO3) and Sodium carbonate (Na2CO3), we do not know the relation. Then we take a sample of 5 gram. Then we burn that at 300 Celsius for like ten minutes. Then the following reaction will happen:

2NaHCO3 -> Na2CO3 + CO2 + H2O

Then we measure the weight of the sample after we have burned it and we get 3.71 gram. Then we need to calculate many gram Sodium Carbonate we started out with, the number of gram in the samble. And that I do not know how to.

I do hope you can follow if not then please tell me and I will try to make it more understandable.
 
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1. This question belongs in the Homework & Coursework subforum.

2. We can not respond to with help unless you first show your work. Please read the posting guidelines for the above forum.
 
1. Then please feel free to move it. And I am sorry that I did not put it in the right forum.

2. We know that:

m(NaHCO3) + m1(Na2CO3) =5g
and
m3(Na2CO3)=3,71g
and we know
m1(Na2CO3)+m2(Na2CO3)=m3(Na2CO3)

Where:
m1 is what we have got to start with and the one we should find.
m2 is what is produced by the NaHCO3
m3 is what we have got after the reaction we know that one

Now we have to find m1, I just cannot see how.
 
You have 4 unknowns: m, m1, m2, m3. So you need 4 equations to be able to solve for them. So far, you've got 3 equations...so you need one more.

Using the balanced equation for the conversion of carbonate to bicarbonate (and using their respective molar masses), can you not find a relation between m and m2? This relation will be the required 4th equation.

Alternatively, you could call m1 = x. Then, m = 5 - x. And from your third equation, m2 = 3.71 - x

So you now have everything expressed in terms of one variable: x.

Now if you figure out the last equation (from the stoichiometry), and express that in terms of x as well, you should be able to solve for x.
 
Thread moved to Homework Help, Other Sciences section.
 
Gokul43201 said:
Using the balanced equation for the conversion of carbonate to bicarbonate (and using their respective molar masses), can you not find a relation between m and m2? This relation will be the required 4th equation.

That is excactly my problem I cannot find the link between those two, m2 and m.
 
lo2 said:
That is excactly my problem I cannot find the link between those two, m2 and m.
Try the following steps:

1. Translate the balanced equation to words: "every 2 moles of bicarbonate produces 1 mole of carbonate"

2. Using their molar masses (molecular weights), rewrite that above sentence using weights instead of moles: "every <blah1> grams of bicarbonate produces <blah2> grams of carbonate"

3. Ratio and proportion: Since we know that m grams of bicarbonate produces m2 grams of carbonate, we must have blah1/blah2 = m/m2

That gives you the final equation.
 

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