chemistry
posted by Some kid .
How many milliliters of 0.418 M HCl are needed to react with 52.7 g of CaCO3?
2HCl (aq) + CaCO3 (s) >>> CaCl2 (aq) + CO2 (g) + H20 (l)
I got .00252 mL but its wrong. This is what I did:
52.7 g x (1 mol CaCO3/100.09 CaCO3) x (2 mol HCl/ 1 mol CaCO3)= 1.05 mol HCl
volume= moles/ molarity
volume= 1.05/.418 = .00252 mL

Thanks for showing your work. That makes it easier to find the error.
You're ok to the last step. I would write
L = moles/molarity just so I would remember that the answer is in liters.
Then 1.05/0.418 = 2.52 L or 2520 mL. Check my work for arithmetic errors.