An excess 0.10 mol dm 3 HCl was poured into a big beaker containing 2g

Explanation

In this question, we are given information about the reaction between excess hydrochloric acid (HCl) and limestone, as well as the neutralization of the unreacted acid with potassium carbonate (K\(_2\)CO\(_3\). The question asks us to determine the original volume of the acid. To solve this problem, we first need to write out the balanced chemical equation for the reaction between HCl and limestone: CaCO\(_3\) + 2HCl ? CaCl\(_2\) + CO\(_2\) + H\(_2\)O This equation tells us that one mole of CaCO\(_3\) reacts with two moles of HCl. Since we are given an excess of HCl, we can assume that all of the CaCO\(_3\) has reacted. We are also given the mass of the CaCO\(_3\) (2g) and its molar mass (100 g/mol), so we can calculate the number of moles of CaCO\(_3\) using the formula: moles = mass/molar mass moles of CaCO\(_3\) = 2g/100 g/mol = 0.02 mol Since one mole of CaCO\(_3\) reacts with two moles of HCl, we know that 0.04 moles of HCl were required to react with the CaCO\(_3\). However, we are told that there is unreacted HCl remaining, which requires 25 cm\(^3\) of 0.10 moldm\(^{-3}\) K\(_2\)CO\(_3\) to neutralize it. This information allows us to calculate the number of moles of HCl that were not used in the reaction with CaCO\(_3\): moles of HCl = concentration x volume moles of HCl = 0.10 moldm\(^{-3}\) x 0.025 dm\(^3\) = 0.0025 mol To find the original volume of the acid, we need to add the volume of the unreacted HCl to the volume of HCl that was used in the reaction with CaCO\(_3\): original volume of HCl = (moles of HCl used in reaction + moles of unreacted HCl)/concentration original volume of HCl = (0.04 mol + 0.0025 mol)/0.10 moldm\(^{-3}\) = 0.425 dm\(^3\) = 425 cm\(^3\) Therefore, the correct answer is Option D: 450 cm\(^3\).