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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

  • chemistry -

    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.

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