13 Calcium carbonate reacts with hydrochloric acid: CaCO_(3)(s)+2HCl(aq)arrow CaCl_(2)(aq)+H_(2)O(l)+CO_(2)(g) Calcium carbonate is added to 25.0cm^3 of 2.00mol/dm^3 hydrochloric acid. (A_(r):C=12,O=16,Ca=40) Calculate the amount in moles of hydrochloric acid. Calculate the amount in moles of CaCO_(3) that reacts with the acid. Calculate the mass of CaCO_(3) that reacts with the acid. Calculate the volume of CO_(2) (measured at rtp)produced. (Molar volume at rtp is 24000cm^3

Let's go through each part of the problem step by step:<br /><br />### 1. Calculate the amount in moles of hydrochloric acid.<br /><br />Given:<br />- Volume of hydrochloric acid, \( V = 25.0 \, \text{cm}^3 \)<br />- Concentration of hydrochloric acid, \( C = 2.00 \, \text{mol/dm}^3 \)<br /><br />First, convert the volume from cm³ to dm³:<br />\[ 25.0 \, \text{cm}^3 = 0.025 \, \text{dm}^3 \]<br /><br />Now, calculate the amount in moles:<br />\[ \text{Moles of HCl} = C \times V = 2.00 \, \text{mol/dm}^3 \times 0.025 \, \text{dm}^3 = 0.050 \, \text{moles} \]<br /><br />### 2. Calculate the amount in moles of \( CaCO_{3} \) that reacts with the acid.<br /><br />From the balanced chemical equation:<br />\[ CaCO_{3}(s) + 2HCl(aq) \rightarrow CaCl_{2}(aq) + H_{2}O(l) + CO_{2}(g) \]<br /><br />The stoichiometry shows that 1 mole of \( CaCO_{3} \) reacts with 2 moles of HCl.<br /><br />Given that we have 0.050 moles of HCl, the moles of \( CaCO_{3} \) that can react are:<br />\[ \text{Moles of } CaCO_{3} = \frac{0.050 \, \text{moles HCl}}{2} = 0.025 \, \text{moles} \]<br /><br />### 3. Calculate the mass of \( CaCO_{3} \) that reacts with the acid.<br /><br />Given:<br />- Molar mass of \( CaCO_{3} \):<br /> - \( Ca = 40 \, \text{g/mol} \)<br /> - \( C = 12 \, \text{g/mol} \)<br /> - \( O = 16 \, \text{g/mol} \)<br /><br />\[ \text{Molar mass of } CaCO_{3} = 40 + 12 + (3 \times 16) = 100 \, \text{g/mol} \]<br /><br />Now, calculate the mass:<br />\[ \text{Mass of } CaCO_{3} = \text{Moles of } CaCO_{3} \times \text{Molar mass of } CaCO_{3} \]<br />\[ \text{Mass of } CaCO_{3} = 0.025 \, \text{moles} \times 100 \, \text{g/mol} = 2.50 \, \text{g} \]<br /><br />### 4. Calculate the volume of \( CO_{2} \) produced (measured at rtp).<br /><br />From the balanced chemical equation, 1 mole of \( CaCO_{3} \) produces 1 mole of \( CO_{2} \).<br /><br />Given that we have 0.025 moles of \( CaCO_{3} \), the moles of \( CO_{2} \) produced are:<br />\[ \text{Moles of } CO_{2} = 0.025 \, \text{moles} \]<br /><br />Given:<br />- Molar volume of \( CO_{2} \) at rtp = 24000 cm³/mol<br /><br />Now, calculate the volume:<br />\[ \text{Volume of } CO_{2} = \text{Moles of } CO_{2} \times \text{Molar volume of } CO_{2} \]<br />\[ \text{Volume of } CO_{2} = 0.025 \, \text{moles} \times 24000 \, \text{cm}^3/\text{mol} = 600 \, \text{cm}^3 \]<br /><br />### Summary of Results:<br />1. Amount in moles of hydrochloric acid: \( 0.050 \, \text{moles} \)<br />2. Amount in moles of \( CaCO_{3} \) that reacts with the acid: \( 0.025 \, \text{moles} \)<br />3. Mass of \( CaCO_{3} \) that reacts with the acid: \( 2.50 \, \text{g} \)<br />4. Volume of \( CO_{2} \) produced: \( 600 \, \text{cm}^3 \)