Determining volume of an acid needed to react with a mass

[Deluxe489]

1. Determine the volume of 1.50 mol dm–3 of hydrochloric acid that would react with exactly 1.25 g of calcium carbonate.



2. 2HCl(aq) + CaCO3(s) --> CaCl2(aq) + CO2(g) + H2O(1)



3. I tried to do this: (1.25 g CaCO3) / (100 g CaCO3) = 0.0125 moles of CaCO3
2 moles of HCl are needed for every 1 of CaCO3; therefore, 2*0.0125 = 0.0250 moles of HCl
(0.0250 moles of HCl)(36.46 g HCl) / (1.18 g HCl) = 7.73 cm3 HCl
Unfortunately, the correct answer is 16.7 cm 3 HCl. I know it has something to do with the 1.50 mol dm-3, but I don't know how to do the problem.

Thank you for any help.

 

[danago]

Your reasoning up to the point where you conclude that you need 0.025 moles of HCl is good. I am not exactly sure what you have done from there (where did 1.18g come from??)

From the definition of molar concentration, c = n/V, you can find the volume V that you require. You have found that n=0.025 moles of HCl and you have the concentration of HCl, so you can find the volume.

 

[Deluxe489]

Thank you. The density of hydrochloric acid is 1.18 g/cm^3, and I thought I needed to use that. I guess I didn't.

 

[Borek]

The density of hydrochloric acid is 1.18 g/cm^3

This is not a density of hydrochloric acid, this is density of a hydrochloric acid SOLUTION.