[Stephen]

1.If we get in 200g NaHSO_4(aq), 200g of Na₂CO_3(aq) then mass of produced solution will be 395.6g, if we do the same thing just in different timetable (In Na₂CO₃ we put NaHSO₄) then mass of produced solution will be 397.8. What are mass percentages of start solutions ? % of Na₂CO₃ and % of  NaHSO₄ ?
I don't have idea for solving this :S
 Na₂CO₃+NaHSO₄  sholdn't give anything right?So, where does this 4.4 and 2.2 grams go?!

[Dan]

Another nice question!

Na₂CO₃+NaHSO₄  sholdn't give anything right?So, where does this 4.4 and 2.2 grams go?!

No, there are some acid-base reactions that can happen - the order of addition is critical to the equilibria you get!
The mass you lose escapes as a gas....

Don't have time to explain fully now, have a think about it - really good question!

Hint: The limiting reagent in each case depends on the order of addition

[Stephen]

Maybe like this:
Na₂CO₃+2NaHSO₄-->2Na₂SO₄+H₂O+CO₂ 
Hmm I started like this, tell me if it's good start
400-395.6=4.4g this should be mass of CO₂ so If 120g(Mr (NaHSO4)) gives 44 grams of CO₂ then x gives 4.4 grams and This means that x=12 that means that there was 12 grams of NaHSO4 and at the same way we can calculate mass of NaCO₃ y=10.6 grams
Well can you tell me am I suppose to product this 120 with 2 because there are 2NaHSO4 in eq. ?Am I suppose to do this when I do Stoichiometric calculation?
So if I product with 2 then x=24, y=10.6 if not than it's x=12 and y=10.6
But I don't see how to find limiting reactant!?
Lets go on with exercise and see what're we going to get:
Now difference of masses is 400-397.8=2.2 g of CO2
So:
x=6 y=5.3 if we don't product with 2 and If we do hen x=12 and y=5.3 
I don't see next step really
Trying to figure out who is limitting reactant by using logic If there is more CO2 ,,lost,, when we add Na2CO3 that has to mean that than more Na2CO3 is used in reaction so assuming that I think that limiting reactant is Na2CO3 that means that in excess is NaHSO4 am I right?! 

[Dan]

OK, I would work in moles rather than grams and then convert back at the end, it's less confusing (at least for me). Consider the following reactions:

NaHSO₄ + Na₂CO₃ <------> Na₂SO₄ + NaHCO₃ [equation1]

NaHSO₄ + NaHCO₃ -----> Na₂SO₄ + CO₂ + H₂O [equation 2]

Conceptually, if you have a solution of sodium hydrogen sulfate and add sodium carbonate to it then the sulfate is always in excess (until it has all reacted). Equation 2 will only happen if there is hydrogen sulfate present in the solution (ie acidic conditions). Under these conditions you will form the maximum amount carbon dioxide, 4.4 g or 0.1 mol.

If you do it the other way round, and add the hydrogen sulfate to the carbonate (basic conditions), only equation 1 will be happening until all of the sodium carbonate has been protonated (changed into hydrogen carbonate). Once you have added enough of the hydrogen sulfate then equation 2 will start happening and you generate 2.2g or 0.05 mol of carbon dioxide.