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Topic: Gas Problem  (Read 12549 times)

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Offline amiv14

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Gas Problem
« 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

« Last Edit: December 11, 2006, 09:45:50 AM by amiv14 »

Offline Yggdrasil

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Re: Need help on question
« Reply #1 on: December 11, 2006, 01:12:48 AM »
Please read the Forum Rules.  You have to show your attempt at answering the question before we can help you.

Offline amiv14

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Re: Need help on question
« Reply #2 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.
« Last Edit: December 11, 2006, 09:47:49 AM by amiv14 »

Online Borek

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Re: Gas Problem
« Reply #3 on: December 11, 2006, 09:57:51 AM »
KOH & HCl - will they react?

K2CO3 & HCl - will they react?
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Offline amiv14

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Re: Gas Problem
« Reply #4 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

Offline Dan

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Re: Gas Problem
« Reply #5 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.
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Offline amiv14

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Re: Gas Problem
« Reply #6 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

Offline Yggdrasil

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Re: Gas Problem
« Reply #7 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.

Offline amiv14

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Re: Gas Problem
« Reply #8 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
« Last Edit: December 12, 2006, 12:15:54 AM by amiv14 »

Offline Yggdrasil

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Re: Gas Problem
« Reply #9 on: December 12, 2006, 01:17:35 AM »
What's the balanced chemical reaction for the reaction between HCl and NaOH?

Offline amiv14

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Re: Gas Problem
« Reply #10 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
« Last Edit: December 12, 2006, 08:25:11 AM by amiv14 »

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Re: Gas Problem
« Reply #11 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.
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Re: Gas Problem
« Reply #12 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?
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Offline amiv14

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Re: Gas Problem
« Reply #13 on: December 12, 2006, 08:32:30 AM »
No i dont know wat that is?

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