## Stoichiometry - 0.7g of Na2CO3.xH2O were dissolved in water...

1. The problem statement, all variables and given/known data
0.7g of Na2CO3.xH2O were dissolved in water and the volume was made to 100 ml, 20 ml of this solution required 19.8 ml of N/10 HCl for complete neutralization. The value of x ?

2. Relevant equations

Na2CO3 + 2HCl = 2NaCl + H2O + CO2

3. The attempt at a solution

I have taken the Normality of sodium carbonate solution as 0.07 N. Since the solution is dissolved in water and not in any other solution, the number of molecules of water in sodium carbonate should not matter for calculating the normality of the solution. This requires 19.8 ml of 0.1 N HCl. I have tried balancing the equivalents, i.e the equivalent of sodium carbonate is Normality x volume = 0.07 * 20 = 0.14. The equivalents of HCl is 0.1 * 19.8 = 1.98.

Where does the "x" come in ?

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 "I have taken the Normality of sodium carbonate solution as 0.07 N." That's your mistake. The problem statement doesn't tell you that. It says 0.7 GRAMS of Na2CO3.xH2O were dissolved, not 0.7 MOLES. In order to solve this problem, you need to figure out how many moles of Na2CO3.xH2O 0.7g is.

 Quote by pmsrw3 "I have taken the Normality of sodium carbonate solution as 0.07 N." That's your mistake. The problem statement doesn't tell you that. It says 0.7 GRAMS of Na2CO3.xH2O were dissolved, not 0.7 MOLES. In order to solve this problem, you need to figure out how many moles of Na2CO3.xH2O 0.7g is.
OK, so the number of moles of Na2CO3.xH2O is 0.7 / 106 + 18x. What next?

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## Stoichiometry - 0.7g of Na2CO3.xH2O were dissolved in water...

 Quote by Quantum Mind OK, so the number of moles of Na2CO3.xH2O is 0.7 / 106 + 18x. What next?
The number of moles of Na2CO3.xH2O is 0.7/(106 + 18x).

Not a pedantic difference.

Na2CO3 + 2HCl = 2NaCl + H2O + CO2

is telling you one mole of Na2CO3 reacts with how many moles of HCl?

From the problem statement how many moles of HCl have reacted? Therefore how many moles did you have there?

 one mole of HCl is 36.5 g, which means one-half mole of HCl has reacted with the aq. sodium carbonate. Since normally one mole of sodium carbonate reacts with two moles of HCl, in this case it means one mole of Na2CO3 has reacted with half a mole of HCl. This means that Na2CO3 is one fourth of a mole? i.e 26.5 g ? But what we have here is only 0.7 g that too Na2CO3.xH2O. How to relate that to the figure 0.7 / 106 + 18x ?

 Quote by Quantum Mind one mole of HCl is 36.5 g, which means one-half mole of HCl has reacted with the aq. sodium carbonate.
Wrong again, Buffalo Jack! 19.8 is not the amount of HCl that reacted with the carbonate. It's the amount of N/10 HCl solution that reacted with carbonate.

You've got to pay attention to the details. As epenguin says, "Not a pedantic difference."

 I am complete newbie to stoichiometry, so please bear with me. I have been struggling with this problem for a long time but I finally figured out this much, though I don't know if this is correct: Since 0.7 grams are dissolved in 100 ml of water, in 20 ml the weight would be 0.7/5 = 0.14 g. This is neutralized by 19.8 ml of 0.1 N HCl i.e. 1.98 ml is required for N HCl. no. of moles of hydrated sodium carbonate = given mass / mol mass = 0.14 /106 + 18x How do I proceed from here?

 Quote by Quantum Mind This is neutralized by 19.8 ml of 0.1 N HCl i.e. 1.98 ml is required for N HCl.
How many moles of bicarbonate would be neutralized by 1.98 ml of N HCl?

 Quote by pmsrw3 How many moles of bicarbonate would be neutralized by 1.98 ml of N HCl?
I should probably equate 1.98/1000 and 0.14 / 106+18x, though I am not sure. If I solve this, I get approximately 2, which is the right answer. Is my approach right?

 Quote by Quantum Mind I should probably equate 1.98/1000 and 0.14 / 106+18x, though I am not sure. If I solve this, I get approximately 2, which is the right answer. Is my approach right?
Umm, no. If you equate 1.98/1000 and 0.14 / 106+18x, you get x = -2. (That's a minus sign.) Unless your bicarbonate is complexed with anti-matter, that is probably not the answer you're looking for.

Take it one step at a time. How many moles of bicarbonate would be neutralized by 1.98 ml of N HCl?

 Quote by pmsrw3 Umm, no. If you equate 1.98/1000 and 0.14 / 106+18x, you get x = -2. (That's a minus sign.) Unless your bicarbonate is complexed with anti-matter, that is probably not the answer you're looking for. Take it one step at a time. How many moles of bicarbonate would be neutralized by 1.98 ml of N HCl?
I give up. Please tell me how to go about? I have spent too long a time on this one problem.

I know 106 g i.e. one mole of Na2CO3 is neutralized by 73 g, i.e 2 moles of HCl. I don't know how to find out how many moles react with 1.98 ml of HCl. This is my first week at Chem.

 Quote by Quantum Mind I give up.
In that case, I also give up. Maybe someone else will tell you the answer.

 Quote by pmsrw3 In that case, I also give up. Maybe someone else will tell you the answer.
OK, so be it.