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.

Answer:

  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.