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Topic: Gas Problem  (Read 10925 times)

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Offline amiv14

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Re: Gas Problem
« Reply #15 on: December 12, 2006, 09:18:10 AM »
Ok i knew what a limiting reactant was. But i wasnt sure which one was the limiting reactant in this problem. Isnt it the OHnegative. But i just dont know how to do it since it is in mL. I know how to do it in grams

Offline Borek

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Re: Gas Problem
« Reply #16 on: December 12, 2006, 09:51:35 AM »
You don't know how to convert concetration and volume to moles? And how to convert moles to grams? (which is a step I would try to avoid - you better try to express data in moles and convert to grams at the end).
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Offline amiv14

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Re: Gas Problem
« Reply #17 on: December 12, 2006, 09:58:41 AM »
Ok that just confused me. Can you please show me wat i am supposed to do. And how i figure out how much HCl I used. Cuz i dont get it

Offline Yggdrasil

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Re: Gas Problem
« Reply #18 on: December 12, 2006, 02:49:26 PM »
Here's an example of how to do your calculation:

You have a solution containing 5mL of an unknown concentration of H2SO4.  To figure out the concentration, you add 0.1M NaOH until the solution is fully neutralized.  You find that it takes 15.8mL of your NaOH solution to neutralize the H2SO4.  How many moles of H2SO4 were in your unknown solution?


To find the number of moles of NaOH we used, we recall the definition of molarity.  Molarity (M) = moles/L.  Therefore, to find the number of moles of NaOH in 15.8mL of 0.1M solution of NaOH, we multiply the concentration by the volume:

0.1 moles/L * 0.0158L = 0.00158 moles

So, in our reaction we used 0.00158 moles of NaOH.  Now, the reaction which you track is the following:

2NaOH + H2SO4 --> 2H2O + Na2SO4

From the balanced chemical reaction, we see that NaOH and H2SO4 react at a 2:1 ratio.  Therefore, when the reaction is complete, the number of moles of NaOH reacted = (1/2) * the number of moles of H2SO4 reacted.  So, we must have started with (1/2)* 0.00158 = 0.00079 moles of H2SO4

Offline Donaldson Tan

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Re: Gas Problem
« Reply #19 on: December 12, 2006, 08:31:37 PM »
Total HCl Added = moles of HCl in 0.100 L of 2.00 M HCl solution = HCl reacted with KOH and K2CO3 + Excess HCl

b. The excess HCl is found to be fully neutralized by 86.6 mL of a 1.50 M NaOH solution.

Moles of Excess HCl = Moles of NaOH neutralised = moles of NaOH in 86.6 mL of a 1.50 M NaOH solution

HCl reacted with KOH and K2CO3 = Total HCl - Excess HCl

Moles of HCl reacted with K2CO3 = (2/1) * Moles of CO2 evolved.
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Offline amiv14

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Re: Gas Problem
« Reply #20 on: December 12, 2006, 11:29:51 PM »
ok i think i finally got the answers. Can someone tell me if these are right? 56.2% is KOH 16.1% is KCl and 27.7% is K2CO3

Offline Yggdrasil

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Re: Gas Problem
« Reply #21 on: December 13, 2006, 12:48:18 AM »
Those are the numbers I got.  Good job.  :)

Offline amiv14

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Re: Gas Problem
« Reply #22 on: December 13, 2006, 08:09:23 AM »
Thank You. Thanx all you guyz that helped

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