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.
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