Final answer:
To react completely with 120 mL of 0.03 M HCl, 0.180162 g of calcium carbonate is required. This is calculated by first determining the moles of HCl present and then using stoichiometry to find the corresponding moles and then the mass of calcium carbonate required.
Step-by-step explanation:
The question pertains to a stoichiometric calculation in chemistry involving a reaction between calcium carbonate (CaCO3) and hydrochloric acid (HCl). Specifically, we want to calculate the mass of calcium carbonate needed to react completely with 120 mL of a 0.03 M HCl solution. The balanced chemical equation for the reaction between calcium carbonate and hydrochloric acid is: CaCO3 + 2 HCl → CaCl2 + CO2 + H2O.
First, we calculate the number of moles of HCl in 120 mL of the solution: 0.03 moles/L × 0.120 L = 0.0036 moles. The stoichiometry of the reaction tells us one mole of CaCO3 reacts with two moles of HCl, thus we need 0.0036 moles / 2 = 0.0018 moles of calcium carbonate. The molar mass of calcium carbonate is 100.09 g/mol, so the mass needed is: 0.0018 moles × 100.09 g/mol = 0.180162 g. Therefore, 0.180162 g of calcium carbonate is needed to react completely with the given volume of HCl solution.