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Chemistry Forums for Students => High School Chemistry Forum => Topic started by: amiv14 on December 11, 2006, 12:39:31 AM

Title: Gas Problem
Post by: amiv14 on December 11, 2006, 12:39:31 AM
A 5.00 g sample of a dry mixture of potassium hydroxide, potassium carbonate, and potassium chloride is reacted with 0.100 L of a 2.00 M HCl solution.

a. A 249 mL sample of dry CO2 gas, measured at 22 degrees Celsius and 740 Torr, is obtained from the reaction. What is the percentage of potassium carbonate in the original mixture?

b. The excess HCl is found to be fully neutralized by 86.6 mL of a 1.50 M NaOH solution. Calculate the percentages of potassium hydroxide and potassium chloride in the original mixture.

Please show all work

Title: Re: Need help on question
Post by: Yggdrasil on December 11, 2006, 01:12:48 AM
Please read the Forum Rules (http://www.chemicalforums.com/index.php?page=forumrules).  You have to show your attempt at answering the question before we can help you.
Title: Re: Need help on question
Post by: amiv14 on December 11, 2006, 09:12:51 AM
Well all i can get is the formula KOH + K2Co3+KCl+HCl then i dont know wat that makes. So can u help me with that atleast

Then i know i got to use stoichiometry to find the amount of Potassium Carbonate. Then i can find the percentage

Then i dont know what it means by neutralized. So then i'm not sure how to find the last part.
Title: Re: Gas Problem
Post by: Borek on December 11, 2006, 09:57:51 AM
KOH & HCl - will they react?

K2CO3 & HCl - will they react?
Title: Re: Gas Problem
Post by: amiv14 on December 11, 2006, 06:21:41 PM
Ok well i got A and i think it is 27.7% is that right. And im totally lost on the second part could you show me how to do it or tell me and then i could do it
Title: Re: Gas Problem
Post by: Dan on December 11, 2006, 07:21:08 PM
Answer Borek's questions, he is trying to help you.

27.7% is probably correct. I got 27.6% by mass, but I was not using very accurate values for my quick calculation.
Title: Re: Gas Problem
Post by: amiv14 on December 11, 2006, 10:37:32 PM
Ok well can anybody help me with part B cuz im really confused. You dont have to do it but please give me instructions or something
Title: Re: Gas Problem
Post by: Yggdrasil on December 12, 2006, 12:05:35 AM
Quote
The excess HCl is found to be fully neutralized by 86.6 mL of a 1.50 M NaOH solution

From this you can figure out how much HCl is left over from the reaction.  Using the original amount of HCl added, you can figure out how much HCl reacted.  Since the HCl reacts with only the potassium carbonate and the potassium hydroxide and you know how much potassium carbonate is in the mixture, you can figure out how much potassium hydroxide there was.
Title: Re: Gas Problem
Post by: amiv14 on December 12, 2006, 12:10:29 AM
How do u find out how much is left over from that. That is the part i dont get. Cuz isnt that pretty much the hard part. I think i might be able to get it from there. Please explain a little more if possible. Or use some of the numbers cuz im not getting it
Title: Re: Gas Problem
Post by: Yggdrasil on December 12, 2006, 01:17:35 AM
What's the balanced chemical reaction for the reaction between HCl and NaOH?
Title: Re: Gas Problem
Post by: amiv14 on December 12, 2006, 08:11:38 AM
Wouldnt it be HCl + NaOH = H0+ NaCl. But if you take out the spectators isnt it just H+OH=H20. So what do u do with that. and how do i find the amount of HCL left over
Title: Re: Gas Problem
Post by: Borek on December 12, 2006, 08:28:53 AM
Please, follow our hints, writing and balancing reaction equations you were asked for. Once the bricks will be ready, we will build a wall.
Title: Re: Gas Problem
Post by: Borek on December 12, 2006, 08:30:12 AM
isnt it just H+OH=H20.

Good (well, almost good, as you have missed the charges).

Now, do you know what limiting reagent is?
Title: Re: Gas Problem
Post by: amiv14 on December 12, 2006, 08:32:30 AM
No i dont know wat that is?
Title: Re: Gas Problem
Post by: Borek on December 12, 2006, 08:56:45 AM
http://www.chembuddy.com/?left=balancing-stoichiometry&right=limiting-reagents

http://en.wikipedia.org/wiki/Limiting_reagent
Title: Re: Gas Problem
Post by: amiv14 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
Title: Re: Gas Problem
Post by: Borek 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).
Title: Re: Gas Problem
Post by: amiv14 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
Title: Re: Gas Problem
Post by: Yggdrasil 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
Title: Re: Gas Problem
Post by: Donaldson Tan 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.
Title: Re: Gas Problem
Post by: amiv14 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
Title: Re: Gas Problem
Post by: Yggdrasil on December 13, 2006, 12:48:18 AM
Those are the numbers I got.  Good job.  :)
Title: Re: Gas Problem
Post by: amiv14 on December 13, 2006, 08:09:23 AM
Thank You. Thanx all you guyz that helped