Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: amiv14 on December 11, 2006, 12:39:31 AM
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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
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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.
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KOH & HCl - will they react?
K2CO3 & HCl - will they react?
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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
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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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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
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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.
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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
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What's the balanced chemical reaction for the reaction between HCl and NaOH?
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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
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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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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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No i dont know wat that is?
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http://www.chembuddy.com/?left=balancing-stoichiometry&right=limiting-reagents
http://en.wikipedia.org/wiki/Limiting_reagent
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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
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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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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
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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
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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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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
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Those are the numbers I got. Good job. :)
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Thank You. Thanx all you guyz that helped