[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%
For some reason I keep getting the wrong answer.

5.93 grams= the mass of NaCl and Na2CO3(s) I beleive

and so 7.78-5.93= 1.85 grams of gas that is released. Because H2O and CO2 are in a 1:1 ratio 0.925 g is lost my each gas.( **side question if they were not 1:1 say 2H2O and 3 CO2..what would you divide each by to get individual mass of each gas)

so my thinking is.. mass of H20-->mols H20--->mol NaHCO3[/sub---> Mass NaHCO3
and than I would subtract the mass of NaHCO3 from 7.78 grams to get mass of NaCl in sample.

So I get 0.05135 mol of H2O *( 2 mol NaHCO3[/sub/1 mol H20)= 0.1027 mol NaHCO3

Molar mass of NaHCO3 =84.007g so multiply this by .1027 and you get 8.63g of NaHCO3
7.78-8.63g = negative mass of NaCl which makes no sence

[Borek]

1.85 grams of gas that is released. Because H2O and CO2 are in a 1:1 ratio 0.925 g is lost my each gas.

1:1 is molar ratio, not mass ratio.