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August 2, 2015

Homework Help: chemistry

Posted by Some kid on Saturday, September 19, 2009 at 4:51pm.

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

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