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 $CaCO_{3}$ in the sample need to follow these steps:<br /><br />1. Calculate the moles of hydrochloric acid (HCl) used in the reaction.<br />2. Calculate the moles of calcium carbonate (CaCO3) that reacted with HCl.<br />3. Calculate the mass of calcium carbonate (CaCO3) in the sample.<br />4. Calculate the \% W/W of calcium carbonate (CaCO3) in the sample.<br /><br />Step 1: Calculate the moles of hydrochloric acid (HCl) used in the reaction.<br />Given:<br />Volume of HCl solution = 20.00 mL<br />Molarity of HCl solution = 0.2254 M<br /><br />Moles of HCl = Molarity × Volume (in liters)<br />Moles of HCl = 0.2254 M × 0.020 L<br />Moles of HCl = 0.004508 moles<br /><br />Step 2: Calculate the moles of calcium carbonate (CaCO3) that reacted with HCl.<br />The balanced chemical equation for the reaction between calcium carbonate (CaCO3) and hydrochloric acid (HCl) is:<br />CaCO3 + 2HCl → CaCl2 + H2O + CO2<br /><br />From 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:<br />Moles of CaCO3 = Moles of HCl / 2<br />Moles of CaCO3 = 0.004508 moles / 2<br />Moles of CaCO3 = 0.002254 moles<br /><br />Step 3: Calculate the mass of calcium carbonate (CaCO3) in the sample.<br />Given:<br />Mass of the calcium carbonate sample = 0.2719 g<br /><br />Molar mass of CaCO3 = 100.09 g/mol<br /><br />Mass of CaCO3 = Moles of CaCO3 × Molar mass of CaCO3<br />Mass of CaCO3 = 0.002254 moles × 100.09 g/mol<br />Mass of CaCO3 = 0.2259 g<br /><br />Step 4: Calculate the \% W/W of calcium carbonate (CaCO3) in the sample.<br />\% W/W of CaCO3 = (Mass of CaCO3 / Mass of the sample) × 100<br />\% W/W of CaCO3 = (0.2259 g / 0.2719 g) × 100<br />\% W/W of CaCO3 ≈ 83.34%<br /><br />Therefore, the \% W/W of calcium carbonate (CaCO3) in the sample is approximately 83.34%.