Chemistry Question, help very much appreciated?

J

jakehartley123

New member
This is the background information i'm given;

Sodium carbonate forms a number of hydrates of general formula Na2CO3.xH20. A 3.01g sample of one of these hydrates was dissolved in water and the solution made up to 250cm^3.
In a titration, a 25cm^3 portion of this solution required 24.3cm^3 of 0.2 mol dm^-3 hydrochloric acid for complete reaction.
The equation for this reaction is shown below;

Na2CO3 + 2HCl = 2NaCl + H2O + CO2

Right, now if you've take the time to read this far, thankyou for caring, now the important part, the questions.

1) Calculate the number of moles of HCl in 24.3 cm^3 of 0.2 mol dm^-3 hydrochloric acid.

2) Deduce the number of moles of Na2CO3 in 25 cm^3 of the Na2CO3 solution.

3) Hence deduce the number of moles of Na2CO3 in the original 250cm^3 of solution


Thankyou very much for reading this far, an I hope you can help me as i'm fully stumped here. Any plausible answers are appreciated, and if some kind person could explain how they reached their answers i'd be more than happy to give them 10 points and my undying love :D

Thankyou xx
 
Being male, I can do without your undying love. The hangup I see is that 02.mole/dm^3 is the same as 0.1 molar. So you can compute the moles of HCl as 0.2 x 24.3 x 10^-3 = 0.0486 moles.

Now, a 1/10th sample would have 0.301 grams of hydrated Na2CO3. BUT whatever the number of waters in each hydrate molecule, each mole of the bicarbonate requires 2 moles of HCl to react. So, the sample would have 0.0243 moles of pure bicarbonate.

That gets us to (3). The answer here is 0.243 moles from the arguement above. What we haven't done (and it wasn't asked for) is the number of waters in each hydrate. If you need help with THIS, try us again.
 
Being male, I can do without your undying love. The hangup I see is that 02.mole/dm^3 is the same as 0.1 molar. So you can compute the moles of HCl as 0.2 x 24.3 x 10^-3 = 0.0486 moles.

Now, a 1/10th sample would have 0.301 grams of hydrated Na2CO3. BUT whatever the number of waters in each hydrate molecule, each mole of the bicarbonate requires 2 moles of HCl to react. So, the sample would have 0.0243 moles of pure bicarbonate.

That gets us to (3). The answer here is 0.243 moles from the arguement above. What we haven't done (and it wasn't asked for) is the number of waters in each hydrate. If you need help with THIS, try us again.
 
Being male, I can do without your undying love. The hangup I see is that 02.mole/dm^3 is the same as 0.1 molar. So you can compute the moles of HCl as 0.2 x 24.3 x 10^-3 = 0.0486 moles.

Now, a 1/10th sample would have 0.301 grams of hydrated Na2CO3. BUT whatever the number of waters in each hydrate molecule, each mole of the bicarbonate requires 2 moles of HCl to react. So, the sample would have 0.0243 moles of pure bicarbonate.

That gets us to (3). The answer here is 0.243 moles from the arguement above. What we haven't done (and it wasn't asked for) is the number of waters in each hydrate. If you need help with THIS, try us again.
 
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