[narutodemonkill]

7.78 g sample of a mixture Nacl and NahCO3 is heated and NaHco3 decomposes into 2 NaHCO3(s)--> Na2Co3(s) + H20(g) + co2(g)
NaCl is stable and does not react, after decomposition sample weights 5.93 g calculate percent by mass of NaCl in original mixture
answer is 35.2%
I have not seen this type of question before so I am not sure how to start. I think we need to find mass of of Na2co3+h20 +co2? and subtract this from 5.93g so we can get the mass of NaCl than take this and divide it bu 7.78 g x100%

but How would we go about finding the masses etc.

[UG]

Hi narutodemonkill,
From the equation 2 NaHCO₃(s)--> Na₂CO₃(s) + H₂O(g) + CO₂(g)
It can be seen than n(H₂O) = n(CO₂)
You know that the change in mass is 7.78 - 5.93. This is the mass of the water and carbon dioxide that has been lost, forget about the Na₂CO₃, you don't need it. The mass is equal to n(amount of substance) x M(Molar mass), so the change in mass is equal to....try and write an equation yourself.

[narutodemonkill]

so 7.78- 5.93= 1.85 grams were lost from the sample by h20 and co2
so 5.93 grams = mass of Nacl + Na2Co3
1:1 ratio so amount of mass lost by each molecule is 1.85/2= .925 g .. .925/(molar mass of water)=925/16+2(1.01)=0.0513 mols of water x 2 mol NaHC03(s)/1 mol water
.1026 mol Nahco3(s) x molar mass of NaHCO3 = mass of NaHco3(s), so than  do 5.93 grams- mass of NaHco3(s) this gives you mass of NaCl so just divide this by 7.78 grams and x 100% to get percent NaCl.
right?

[UG]

so 7.78- 5.93= 1.85 grams were lost from the sample by h20 and co2
so 5.93 grams = mass of Nacl + Na2Co3
1:1 ratio so amount of mass lost by each molecule is 1.85/2= .925 g .. .925/(molar mass of water)=925/16+2(1.01)=0.0513 mols of water x 2 mol NaHC03(s)/1 mol water
.1026 mol Nahco3(s) x molar mass of NaHCO3 = mass of NaHco3(s), so than  do 5.93 grams- mass of NaHco3(s) this gives you mass of NaCl so just divide this by 7.78 grams and x 100% to get percent NaCl.
right?

I have a bit of trouble following your working here, but what I has hinting at: 1.85 grams = mass of water + mass of carbon dioxide. Since mass = n x M you can write for example, mass of water = n x molar mass. Since nH₂O = nCO₂, all you have to do is put in their molar masses and you can find 'n' with a bit of simple algebra. Whatever this is will be half of n(NaHCO₃), then you can calculate the mass and also the mass of NaCl and hence the % composition.

[Borek]

Nacl NahCO3 NaHco3 NaHCO3 Na2Co3 co2 Na2co3 h20

Especially Na₂Co₃ catched my attention - is it sodium cobalide?

[narutodemonkill]

so 7.78- 5.93= 1.85 grams were lost from the sample by h20 and co2
so 5.93 grams = mass of Nacl + Na2Co3
1:1 ratio so amount of mass lost by each molecule is 1.85/2= .925 g .. .925/(molar mass of water)=925/16+2(1.01)=0.0513 mols of water x 2 mol NaHC03(s)/1 mol water
.1026 mol Nahco3(s) x molar mass of NaHCO3 = mass of NaHco3(s), so than  do 5.93 grams- mass of NaHco3(s) this gives you mass of NaCl so just divide this by 7.78 grams and x 100% to get percent NaCl.
right?

I have a bit of trouble following your working here, but what I has hinting at: 1.85 grams = mass of water + mass of carbon dioxide. Since mass = n x M you can write for example, mass of water = n x molar mass. Since nH₂O = nCO₂, all you have to do is put in their molar masses and you can find 'n' with a bit of simple algebra. Whatever this is will be half of n(NaHCO₃), then you can calculate the mass and also the mass of NaCl and hence the % composition.





yeah sorry I was in a rush but that is what I did.
Thanks