Question

\[{\rm{4}}\,{\rm{gm}}\] of a mixture of \[{\rm{CaC}}{{\rm{O}}_{\rm{3}}}\] and \[{\rm{Si}}{{\rm{O}}_3}\] is treated with \[{\rm{HCl}}\] and \[0.88\,{\rm{gm}}\] of \[{\rm{C}}{{\rm{O}}_{\rm{2}}}\] is produced. What is the percentage of \[{\rm{CaC}}{{\rm{O}}_{\rm{3}}}\] in the original mixture?

Answer 1

Hint: As we know that calcium carbonate is dissolved in acid but silicates do not form any product with acid due to its polymeric chain. This question can be calculated if we could know that one mole of \[{\rm{CaC}}{{\rm{O}}_{\rm{3}}}\] produces how much gram of carbon dioxide.

Complete solution
When any reaction occurs in any medium, it totally depends upon the energy evolved or absorbed by the formation of products or breaking of reactants. When any compound is dissolved in water, another form of energy is produced which is known as hydration energy.
Now coming on our given question,
When one mole \[{\rm{(100}}\,{\rm{g)}}\] of \[{\rm{CaC}}{{\rm{O}}_{\rm{3}}}\] is treated with \[{\rm{HCl}}\], then 1 mole of carbon dioxide\[{\rm{(44}}\,{\rm{g)}}\] is formed as
\[
{\rm{CaC}}{{\rm{O}}_{\rm{3}}}\,{\rm{ + }}\,{\rm{2HCl}} \to \,{\rm{C}}{{\rm{O}}_{\rm{2}}}\,{\rm{ + }}\,{\rm{CaC}}{{\rm{l}}_{\rm{2}}}{\rm{ + }}{{\rm{H}}_{\rm{2}}}{\rm{O}}\\
{\rm{100}}\,{\rm{g}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\;\;\;\;\;\;{\rm{44}}\,{\rm{g}}
\]
When \[{\rm{Si}}{{\rm{O}}_3}\]is treated with \[{\rm{HCl}}\], no reaction occurs, which means that \[{\rm{C}}{{\rm{O}}_{\rm{2}}}\] is produced only by \[{\rm{CaC}}{{\rm{O}}_{\rm{3}}}\].
So, we are given as that \[0.88\,{\rm{gm}}\] of \[{\rm{C}}{{\rm{O}}_{\rm{2}}}\] is formed
Then,\[{\rm{44}}\,{\rm{gm }}\] of \[{\rm{C}}{{\rm{O}}_{\rm{2}}}\] is produced by \[{\rm{100}}\,{\rm{gm}}\] of \[{\rm{CaC}}{{\rm{O}}_{\rm{3}}}\] then
\[0.88\,{\rm{gm}}\]of \[{\rm{C}}{{\rm{O}}_{\rm{2}}}\]is formed by \[{\rm{ = }}\,\dfrac{{{\rm{100}}\,{\rm{gm}}}}{{{\rm{44}}\,{\rm{gm}}}}{\rm{ \times }}\,{\rm{0}}{\rm{.88}}\,{\rm{gm}}\] \[{\rm{CaC}}{{\rm{O}}_{\rm{3}}}\]
Therefore, \[{\rm{2}}\,{\rm{gm}}\] of \[{\rm{CaC}}{{\rm{O}}_{\rm{3}}}\] produces \[0.88\,{\rm{gm}}\] of \[{\rm{C}}{{\rm{O}}_{\rm{2}}}\].
The percentage of original mixture of \[{\rm{CaC}}{{\rm{O}}_{\rm{3}}}\]is calculated by the formula as
\[{\rm{\% }}\,{\rm{of CaC}}{{\rm{O}}_{\rm{3}}}{\rm{ = }}\,\dfrac{{{\rm{calculated}}\,{\rm{weight}}}}{{{\rm{total}}\,{\rm{given}}\,{\rm{weight}}\,{\rm{of}}\,{\rm{mixture}}\,}}{\rm{ \times }}\,{\rm{100}}\]
Putting the calculated weight and given weight we get our answer as-
\[
{\rm{\% }}\,{\rm{of CaC}}{{\rm{O}}_{\rm{3}}}{\rm{ = }}\,\dfrac{{{\rm{2}}\,{\rm{gm}}}}{{{\rm{4}}\,{\rm{gm}}\,}}{\rm{ \times }}\,{\rm{100}}\\
{\rm{ = }}\,{\rm{50}}\,{\rm{\% }}
\]

Therefore, our final answer is percentage of \[{\rm{CaC}}{{\rm{O}}_{\rm{3}}}\] in the original mixture is \[{\rm{50}}\,{\rm{\% }}\].

Note:
Calcium carbonates when dissolves in water, it is very less soluble because the lattice energy of calcium carbonate is greater than hydration energy but when it dissolves in acid, it is soluble and releases carbon dioxide gas.