A sample of calcium carbonate is dissolved with 20.00 mL . of 0.22 :54 M hydrochloric acid and the excess acid is titrated with 0.10141 M sodium hydroxide. After dissolution, a mass of 0.2719 g of the calcium carbonate sample requires a titre of 12.39 mL of sodium hydroxide to reach a phenolphthalein end point. Find I the % W/W of CaCO_(3) in the sample.

To find the \% W/W of in the sample need to follow these steps:1. Calculate the moles of hydrochloric acid (HCl) used in the reaction.2. Calculate the moles of calcium carbonate (CaCO3) that reacted with HCl.3. Calculate the mass of calcium carbonate (CaCO3) in the sample.4. Calculate the \% W/W of calcium carbonate (CaCO3) in the sample.Step 1: Calculate the moles of hydrochloric acid (HCl) used in the reaction.Given:Volume of HCl solution = 20.00 mLMolarity of HCl solution = 0.2254 MMoles of HCl = Molarity × Volume (in liters)Moles of HCl = 0.2254 M × 0.020 LMoles of HCl = 0.004508 molesStep 2: Calculate the moles of calcium carbonate (CaCO3) that reacted with HCl.The balanced chemical equation for the reaction between calcium carbonate (CaCO3) and hydrochloric acid (HCl) is:CaCO3 + 2HCl → CaCl2 + H2O + CO2From the balanced equation, we can see that 1 mole of CaCO3 reacts with 2 moles of HCl. Therefore, the moles of CaCO3 that reacted with HCl can be calculated as:Moles of CaCO3 = Moles of HCl / 2Moles of CaCO3 = 0.004508 moles / 2Moles of CaCO3 = 0.002254 molesStep 3: Calculate the mass of calcium carbonate (CaCO3) in the sample.Given:Mass of the calcium carbonate sample = 0.2719 gMolar mass of CaCO3 = 100.09 g/molMass of CaCO3 = Moles of CaCO3 × Molar mass of CaCO3Mass of CaCO3 = 0.002254 moles × 100.09 g/molMass of CaCO3 = 0.2259 gStep 4: Calculate the \% W/W of calcium carbonate (CaCO3) in the sample.\% W/W of CaCO3 = (Mass of CaCO3 / Mass of the sample) × 100\% W/W of CaCO3 = (0.2259 g / 0.2719 g) × 100\% W/W of CaCO3 ≈ 83.34%Therefore, the \% W/W of calcium carbonate (CaCO3) in the sample is approximately 83.34%.