[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---> 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 NaHCO₃/1 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

[sjb]

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

Agreed

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.

Not agreed. If you had 1 mole of water, what mass would that be? And 1 mole of carbon dioxide? Is the mass ratio 1:1 in this case?

[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

Agreed

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.

Not agreed. If you had 1 mole of water, what mass would that be? And 1 mole of carbon dioxide? Is the mass ratio 1:1 in this case?

it would be molar mass so 20.16 gram H20 and 44.01 grams CO2
Coefficient would not change outcome?

1 : 2.18  ratio? I'm not sure what you are suppose to do from this

20.16/64.17=.31  and 44.01/64.17=0.686
.......20.16 is molar mass of H2O and 44.01 is molar mass of CO2 64.17 is there sum.
.31 *1.85=.5735 gram h20 and .686*1.85=1.2691 gram co2

if this is correct... how would the amounts change if there were different coefficients.
say 2h2o and 1 co2
instead would you just multiply the molar mass of h20 by 2 in this case and do the same process as above to get the percentage.

[sjb]

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.

Not agreed. If you had 1 mole of water, what mass would that be? And 1 mole of carbon dioxide? Is the mass ratio 1:1 in this case?

it would be molar mass so 20.16 gram H20 and 44.01 grams CO2
Coefficient would not change outcome?

1 : 2.18  ratio? I'm not sure what you are suppose to do from this

20.16/64.17=.31  and 44.01/64.17=0.686
.......20.16 is molar mass of H2O and 44.01 is molar mass of CO2 64.17 is there sum.
.31 *1.85=.5735 gram h20 and .686*1.85=1.2691 gram co2

if this is correct... how would the amounts change if there were different coefficients.
say 2h2o and 1 co2
instead would you just multiply the molar mass of h20 by 2 in this case and do the same process as above to get the percentage.

Check the molecular mass of water, but yes, if the stoichiometry showed that there were 2 moles of water and 1 of carbon dioxide given off, then in x g of missing mass, there would be y mol of water, and z g of CO₂, where y and z are in the ratio 2:1

[Borek]

there would be y mol of water, and z g of CO₂, where y and z are in the ratio 2:1

I am sure you meant something else, not ratio of moles/mass.

[sjb]

Oops, yes, sorry

for z g read z mol

[HSUMSAM]

2NaHCO3(s)--> Na2CO3(s) + H20(g) + CO2(g)
the mole ratio of CO2 and H2O is 1:1 but not 0.925g each

[narutodemonkill]

2NaHCO3(s)--> Na2CO3(s) + H20(g) + CO2(g)
the mole ratio of CO2 and H2O is 1:1 but not 0.925g each

yes it is 1:1 mol ratio not Mass, yet I still do not know how to answer question.

[Borek]

Don't you know how to convert molar ratio to mass ratio?