moles of chalk lab key
d: For every mole of CaCO₃, 2 moles of HCl are needed. By calculating the moles of chalk, you can adjust the volume or concentration of acid accordingly. Example Calculation If you have 0.05 mol of chalk: Moles of HCl needed = 2 × 0.05 mol = 0.10 mol For a 1 M HCl solution, volume needed =