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Topic: Calculate the percentage....  (Read 8368 times)

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ronnie

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Calculate the percentage....
« on: April 22, 2006, 08:59:27 AM »
Would you please answer the following Q ?

0.949 g of mixture NaOH & NaCO3 dissolve in 100 cm3 Water. Then 30cm3, 0.65M HCL added into it for complete neutralization.
Calculate the percentage of NaOH, NaCO3 respectively ?

Please show the steps ~

thanks ~

Offline Borek

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Re: Calculate the percentage....
« Reply #1 on: April 22, 2006, 09:34:00 AM »
You have to set up a system of two equations - one will be mass + mass, second will be number of moles + number of moles. You have two unknowns - be it masses of both substances - that can be inserted into these two equations (hint: you will need their molar masses).

You also better read forum rules:

http://www.chemicalforums.com/index.php?topic=59.0
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Offline Qazzian

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Re: Calculate the percentage....
« Reply #2 on: April 22, 2006, 09:38:07 AM »
Should that be NaHCO3, or Na2CO3?
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Offline Borek

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Re: Calculate the percentage....
« Reply #3 on: April 22, 2006, 10:19:12 AM »
Should that be NaHCO3, or Na2CO3?

Does it matter?
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Offline Qazzian

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Re: Calculate the percentage....
« Reply #4 on: April 22, 2006, 11:20:48 AM »
Wouldn't it matter for the number of moles HCl to neutrilize it?
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Offline Albert

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Re: Calculate the percentage....
« Reply #5 on: April 22, 2006, 11:38:16 AM »
There's nothing wrong in the opening post. Except for the lack of attempts.

Offline xiankai

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Re: Calculate the percentage....
« Reply #6 on: April 22, 2006, 12:09:09 PM »
even if the Na2CO3 became NaHCO3, amt of HCl used will be the same;

Na2CO3 = 2NaHCO3 = 2HCl

unless the water happens to acidify the sodium carbonate... maybe?
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Offline Borek

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Re: Calculate the percentage....
« Reply #7 on: April 22, 2006, 12:14:22 PM »
Wouldn't it matter for the number of moles HCl to neutrilize it?

What I mean is - it matters for the final result, but it doesn't change methodology needed to solve the question.
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Offline Borek

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Re: Calculate the percentage....
« Reply #8 on: April 22, 2006, 12:22:09 PM »
even if the Na2CO3 became NaHCO3, amt of HCl used will be the same;

Na2CO3 = 2NaHCO3 = 2HCl

Think it over - x moles of carbonate requires twice more HCl than x moles of hydrogen carbonate.
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Offline xiankai

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Re: Calculate the percentage....
« Reply #9 on: April 25, 2006, 05:28:32 AM »
oh wait u mean something like this happens?

Na2CO3 + H2O --> NaHCO3 + NaOH

in that case i made a mistake... i was thinking more along the lines of

Na2CO3 + 2H+ --> 2NaHCO3

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Offline Borek

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Re: Calculate the percentage....
« Reply #10 on: April 25, 2006, 10:51:55 AM »
No, you have two reactions taking place (in fact there are three - carbonates react in two steps - but it doesn't matter for the final result):

NaOH + HCl -> NaCl + H2O
Na2CO3 + 2HCl -> 2NaCl + H2O + CO2
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Offline xiankai

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Re: Calculate the percentage....
« Reply #11 on: April 26, 2006, 04:38:29 AM »
oh i understand now... intially i was thinking that the only way that NaHCO3 could be formed was via the decomposition of Na2CO3, but after re-reading the entire thread i realise its all due to a typo error in the first place, thanks Borek!

<resolves to read posts more caerfully in the future>




Edit: missing 2 added
« Last Edit: April 26, 2006, 05:30:00 AM by Borek »
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