posted by Jason .
How many milliliters of 0.237 M HCl are needed to react with 52.2g of CaCO3?
2HCl(aq) + CaCO3(s) --> CaCl2(aq) + CO2(g) + H2O(l)
mol wt of CaCO3 = 40.08 + 12.01 + 48 = 100.09
so, you have 52.2/100.09 = .522 moles
the equation indicates you need 2 moles of HCl for every mole of CaCO3, so you need 1.044 moles HCl
so, now you have to find out how many ml of .237M HCl contain 1.044 moles HCl
Obviously, 1000ml contain .237 moles. So, you need 1.044/.237 * 1000 = 4405.1 ml