• # question_answer 87)   Calcium carbonate reacts with aqueous HCl to give $CaC{{l}_{2}}$ and $C{{O}_{2}}$ according to the reaction given below : $CaC{{O}_{3}}(s)+2HCl(aq)\xrightarrow[{}]{{}}CaC{{l}_{2}}(aq)$$+C{{O}_{2}}(g)+{{H}_{2}}O(l)$     What mass of $CaC{{l}_{2}}$ will be formed when 250 mL of 0.76 M HCl reacts with 1000 g of    $CaC{{O}_{3}}$?  Name the limiting reagent. Calculate the number of moles of $CaC{{l}_{2}}$ formed in the reaction.

Number of moles of HCl taken = $\frac{MV}{1000}=\frac{0.76\times 250}{1000}=0.19$ Number of moles of $CaC{{O}_{3}}=\frac{\text{Mass}}{\text{Molar}\,\text{mass}}=\frac{1000}{100}=10$ The given reaction is: $\underset{1\,mol}{\mathop{CaC{{O}_{3}}}}\,+\underset{2\,mol}{\mathop{2HCl}}\,\to \underset{1\,mol}{\mathop{CaC{{l}_{2}}}}\,+C{{O}_{2}}+{{H}_{2}}O$ Case I:   Let $CaC{{O}_{3}}$ is completely consumed. $1\,mol\,CaC{{O}_{3}}\equiv 1\,mol\,CaC{{l}_{2}}$ $\therefore$   $10mol\,CaC{{O}_{3}}\equiv 10mol\,CaC{{l}_{2}}$ Case ll: Let $HCl$is completely consumed. $2\text{ }mol\text{ }HCl\text{ }=\text{ }1\text{ }mol\text{ }CaC{{l}_{2}}$ $0.19\text{ }mol\text{ }HC1\text{ }\equiv \text{ }\frac{1}{2}\text{ }\times \text{ }0.19\text{ }mol\text{ }CaC{{l}_{2}}\text{ }\equiv 0.095\text{ }mol\text{ }CaC{{l}_{2}}$ Since $HCl$ on complete consumption gives least amount of product hence $HCl$ will be limiting reagent and the number of moles of $CaC{{l}_{2}}$ formed will be 0.095 mol.