How many milliliters of 0.593 M HCl are needed to react with 22.4 g of CaCO3?

Question

2HCl(aq) + CaCO₃(s) > CaCl₂(aq) + CO₂(g) + H₂O(l)

I got 755.48 mL but I know that is wrong.

Answer 1

Hi Dawson! You're actually really close to the answer. Here's the work below:

(0.593 mol / 1L) x (1L / 10³ mL) = (5.93 x 10^-4 mol / 1mL)

22.4g CaCO₃ x (1 mol / 100.087g CaCO₃) x (2 mol HCl / 1 mol CaCO₃) x (1mL / 5.93 x 10^-4 mol)

= 755 mL

Hope this helps! Feel free to reach out if you have any questions or help needed.