Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: TJFCDA on August 03, 2005, 12:06:06 PM
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What is the quantity of 32% HCL required to dissolve one (1) pound of CaCO3 (calcium carbonate)?
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first this question can't be made if you don't know the density of the HCl-solution.
If you know the density of the HCl-solution, then take, for example, 1L solution. You can calculate the mass of 1L HCl with the density and using the mass% you know the mass of HCl in 1L.
--> convert to mol and you know the molarity of the HCl-solution (why?)
convert the mass of CaCO3 to mols and then you can calculate how many mL HCl-solution you need to dissolve 1 pound CaCO3.
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convert 1pound of CaCO3 into number of moles
CaCO3 + 2HCl => CaCl2 + H2O + CO2
every mole of CaCO3 requires 2 moles of HCl to react completely
given the concentration of the HCl solution, find the required volume of the solution to react with 1pound of CaCO3.
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I know that 32% HCL has a density of approximately 10 lbs. per gallon and the molarity is 10.17.
The equation indicates that 1 mole of CaCO3 reacts with 2 moles of HCL. If 1 mole of CaCO3 = 100.0892g and 1 moles of HCL = 36.461g, then it take 72.922g of HCL to dissolve 1 mole of CaCO3. Is this correct?
Assuming this is correct. Shouldn't it require 0.729 lbs of HCL to dissolve 1 lb. of CaCO3?
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first this question can't be made if you don't know the density of the HCl-solution.
Can be done - you just have to give the answer in mass units.
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I know that 32% HCL has a density of approximately 10 lbs. per gallon and the molarity is 10.17.
Doesn't matter.
The equation indicates that 1 mole of CaCO3 reacts with 2 moles of HCL. If 1 mole of CaCO3 = 100.0892g and 1 moles of HCL = 36.461g, then it take 72.922g of HCL to dissolve 1 mole of CaCO3. Is this correct?
Correct.
Assuming this is correct. Shouldn't it require 0.729 lbs of HCL to dissolve 1 lb. of CaCO3?
Yes - but that's not the final answer. This is mass of pure HCl and HCl is only 32% of the solution you have to use.
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Can be done - you just have to give the answer in mass units.
agreed, but when you want to calculate the volume you'll have to know the molarity of the HCl-solution. That's where i was aiming at.
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If I'm using 32% HCL: 0.729/.32=2.27 lbs. of 32% HCL. Is this correct?
Thanks for your assistance.
Troy
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If I'm using 32% HCL: 0.729/.32=2.27 lbs. of 32% HCL. Is this correct?
Seems OK.
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I calculated the amount of HCL using the following method and came up with a slightly different quantity. Is this correct?
1 lb. of CaCO3 = 453.59 g CaCO3
453.29 g CaCO3/100.09 g per mole CaCO3 = 4.532 moles CaCO3
4.532 moles CaCO3 x (2 moles HCL/1 mole CaCO3) = 9.064 moles HCL
9.064 moles HCL x (1 liter solution/10.17 moles HCL) = 0.89 liters of 32% HCL
@ 10 lbs/gallon for 32% HCL
0.98 liters of 32% HCL = 0.235 gallons or 2.35 lbs of 32% HCL
Which of my calculations is the most accurate?
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Which of my calculations is the most accurate?
First one, as it uses only sure numbers.
I know that 32% HCL has a density of approximately 10 lbs. per gallon and the molarity is 10.17
How good is this approximation?
32.0000% HCl solution is 10.1755M and has a density 1.1594 g/mL.
To say the truth, I have no idea what are pound and gallon in metric terms.
I am assuming 0.89/0.98 is a typo.
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Sorry for the typo. Thanks for your support! :)
Troy