Final answer:
To produce 1 gram of CACO₃, we need 100 mL of 0.100 M CACl₂ based on the stoichiometry of the balanced chemical reaction and the given solution concentration.
Step-by-step explanation:
To find out how many milliliters of 0.100 M CACl₂ are needed to produce 1 gram of CACO₃, first, determine the molar mass of CACO₃. Calcium carbonate (CACO₃) has a molar mass of approximately 100.09 g/mol. This means that to get 1 gram of CACO₃, we need 0.01 moles of CACO₃.
According to the balanced equation NACO₃ + CACl₂ → 2NACL + CACO₃, 1 mole of CACl₂ reacts to produce 1 mole of CACO₃. Therefore, we need 0.01 moles of CACl₂ to produce 1 gram of CACO₃.
Since we have a solution of CACl₂ with a concentration of 0.100 M, which means 0.100 moles per liter, we can now calculate the volume needed to get 0.01 moles. To find the volume in liters, we divide the moles of solute by the concentration: Volume (L) = Moles / Concentration. Thus, Volume (L) = 0.01 moles / 0.100 mol/L = 0.1 L.
Since 1 liter is equivalent to 1000 milliliters, we multiply the volume in liters by 1000 to get the volume in milliliters: Volume (mL) = 0.1 L * 1000 mL/L = 100 mL.
Therefore, 100 mL of 0.100 M CACl₂ is required to produce 1 gram of CACO₃.