[TJFCDA]

What is the quantity of 32% HCL required to dissolve one (1) pound of CaCO3 (calcium carbonate)?

[sdekivit]

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.

[Donaldson Tan]

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.

[TJFCDA]

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?

[Borek]

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.

[Borek]

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.

[sdekivit]

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.

[TJFCDA]

If I'm using 32% HCL:  0.729/.32=2.27 lbs. of 32% HCL.  Is this correct?
Thanks for your assistance.
Troy

[Borek]

If I'm using 32% HCL:  0.729/.32=2.27 lbs. of 32% HCL.  Is this correct?

Seems OK.

[TJFCDA]

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?

[Borek]

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.

[TJFCDA]

Sorry for the typo.  Thanks for your support!
Troy