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

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The discussion revolves around calculating the amount of Sodium Carbonate in a sample after combustion of a mixture containing Sodium Bicarbonate and Sodium Carbonate. A 5-gram sample is burned, resulting in a weight of 3.71 grams, prompting the need for calculations to determine the initial amounts of each compound. Participants emphasize the necessity of establishing a fourth equation to solve for the unknowns, using stoichiometric relationships from the balanced chemical reaction. Suggestions include expressing all variables in terms of one unknown and applying molar mass ratios to derive the final equation needed for the calculation. The conversation highlights the importance of following the forum's guidelines for homework help.
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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