Question

How many millimoles of HCl (aq) are neutralized by 0.35 g of CaCO3 (sol)? mmol

Answer 1

it takes two moles of this acid to neutralize one mole of chalk.

millimolesHCl=.35/molmassCaCO3 * 2molHCL/1molCaCO3 *1mol/1000millimol

= .35/100*2*1E-3 check that.