[enantiomorph]

Hi I just have a stoichiometry problem that I couldn't figure out on to solve.  I was hoping if someone could help me out.  Thank you.

Question: Calcium Carbonate (CaCO₃) is put in 250 mL of 0.472M HCl.  All of the CaCO₃ reacts, releasing carbon dioxide gas and leaving behind a clear solution.  20.00mL of the solution was pipetted to another flask.  It took 27.2mL of 0.0562 M NaOH to titrate the HCl remaining in the 20.00 mL portion.  What was the initial mass of CaCO₃?

I'm not sure if the value 20.00 mL play any role in the calculations to solve this problem.  What I had in mind was first calculating the initial numer of moles of HCl = (0.25L)(0.472M)=0.118 mol.  Since NaOH and HCl react in a 1:1 ratio, we can calculate the number of moles of NaOH required to titrate the HCl remaining in the 20.00 mL = (0.0272 L)(0.0562 M) = 0.00153 mol
From writing out the balanced chemical equation of calcium carbonate reacting with hydrochloric acid, we see that the mole ratio of CaCO₃:HCl=1:2
The number of moles of HCl₃ that reacted in the reaction with calcium carbonate = (iniital number of moles) - (moles of NaOH needed for titration) = 0.118 mol - 0.00153 mol = 0.116mol.
The number of moles of CaCO₃ = (1/2)(0.116 mol) = 0.0058 mol
Therefore the mass of CaCO₃ = (0.0058 mol)(40.08+12.01+48)g/mol = 5.82 g

However, I don't understand why this is not the right answer.  Perhaps, somewhere along the way the 20.00 mL of remaining HCl is factored in.  I would greatly appreciate any help with this.  Thanks again.

[Albert]

I think you are on the right track: you take 20 mL out of 250 mL, so, you need to calculate a proportion, when you determine the moles of hydrochloric acid that react with calcium carbonate.

[enantiomorph]

Hi thanks alot Albert!!  You were right, factoring in the 20 mL and calculate the proportion did the trick.

So the solution should resemble something like this:
The balanced chemical equation of the first reaction is CaCO₃ + 2HCl -> CO₂ + H₂O + CaCl₂
The balanced chemical equation of the second reaction is HCl + NaOH -> NaCl + H₂O
The initial amount of moles of HCl = (0.25 L)(0.472 M)=0.118 mol
After the first reaction, 0.25 L of solution remained but some of the initial moles of HCl were consumed.  We took 0.02 L from 0.25 L solution to perform a titration.  We found that it required 0.0272 L of 0.0562 M NaOH to titrate.
From the second balanced chemical equation, we see that HCl:NaOH = 1:1.  Therefore the moles of HCl in 0.02 L of the 0.25 L solution = (0.0272 L)(0.0562 M)=0.00153 mol
By proportion, the number of moles in 0.25 L of the solution that remained following the first reaction is 0.0191 mol.
Thus, the number of moles of HCl that were consumed in the first equation = 0.118 mol - 0.0191 mol = 0.0988 mol
From the first balanced chemical equation we saw that CaCO₃:HCl = 1:2.  Then moles of CaCO₃ = (0.0988mol)/2
Finally, the mass of CaCO₃ = (1/2)(0.0988 mol)(40.08 + 12.01 + 48)g/mol = 4.95 g.

[Albert]

It looks correct to me.